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SAT Formula Sheet new

SAT Formula Sheet

Is memorizing formulas enough to succeed in the SAT exam?

Memorizing formulas for the SAT exam is not enough! 

First of all, you need to understand each formula to decide if it is the right formula needed to solve the question.

Secondly, you need to know how to apply the chosen formula correctly in the context of the question. 

Therefore, you need to learn the formulas and practice to apply them in different questions.    

Why it is the best way to study formulas from this page

In addition to the formulas that are given in the SAT math section we teach many additional formulas that are needed for the SAT exam.

The formulas are divided into groups according to the relevant SAT subscore, so you can start practice solving all subscore questions after you finish learning its formulas.

Learn how and why the formula works, so it will be much easier to remember the formula and use it. 

Each formula is accompanied by a numeric example for a basic understanding of its components. In addition, each formula is accompanied by exercises to test deeper understanding of the formula and make sure you know how to implement the formula correctly and fast.

Before you start learning the formulas

Formula components

Note that formulas include numbers, letters and variables (x, y, z):

  • The numbers have a known value.
  • The letters represent unknown quantities.
  • The variables can change.

The ± symbols in formulas

Note that the ± symbols in formulas have a multiple meaning, so you must choose the top sign (+) or the bottom sign (-).

Once you chose the position of the first symbol, you must continue to choose the same position in all other symbols that follow.

Finding which formula to use

To solve a question, you may need to substitute given numbers for variables and letters to find which formula to use and only then solve with this formula.

For example: Solving the expression 35*21-35*20 requires to use the formula ax-bx =(a-b)x.

ax-bx=35*21-35*20 so that x=35 (this is the value that is the same in both parts of the expression so it must be x), a=21 and b=20.

Since the formula is ax-bx=(a-b)x we get 21*35-20*35=(21-20)*35=1*35=35.

Basic algebra formulas

Distributing formula for removing parentheses: (a±b)x=ax±bx

This is a very basic and known formula. The formula has 2 forms (minus or plus):

(a+b)x=ax+bx

(a-b)x=ax-bx

According to this formula (with a plus sign) instead of adding a and b and then multiplying the sum by x you multiply a by x and then multiply b by x and after that sum the 2 outcomes.

Consider the following examples for using this formula with plus and minus signs:

An example for solving this formula with a plus sign:

There were 5 boxes in the warehouse, each box contained 2 red balls and an unknown number of blue balls. If the total number of balls was 15, that is the number of blue balls in each box?

(2+x)5=15

10+5x=15

5x=5

x=1

Checking the answer: (2+1)5=3*5=15.

An example for solving this formula with a minus sign:

Each box in the warehouse contained 10 products, the worker decided to send less products, so he took an unknown number of products from each box. If the total number of boxes was 5 and the total number of products was 15, that is the number of products the worker took from each box?

(10-x)5=15

50-5x=15

-5x=-35

x=7

Checking the answer: (10-7)5=3*5=15.

Combining like terms formula: ax±bx=(a±b)x

This formula is the same as the previous distributing formula for removing parentheses, you just need to switch the left and the right sides of the formula.

Consider the following examples for using this formula with plus and minus signs: 

An example for solving this formula with a plus sign:

Each table costs 2,000 dollars and each chair costs 500 dollars. If the restaurant needs to purchase 8 tables and there are 6 chairs in each table, what is the total cost for the restaurant?

Note that we can present 2,000 as 500*4.

For the cost of the tables we have the expression (500*4)*8

For the cost of the chairs, we have the expression 500*6*8

To find the total cost we need to solve (500*4)*8 + 500*6*8

We can use the combining like terms formula so that x=500*8 (this are the values that are the same in both parts of the expression so they should be x) getting:

500*4*8 + 500*6*8 = 500*8(4+6) = 500*8*10 = 5,000*8 = 40,000.

Note that we could also define x as x=500*8*2 getting 500*8*2(2+3) = 4,000*2*5 = 8,000*5 = 40,000.

An example for solving this formula with a minus sign:

The cost of one computer is 400 dollars. If there were 20 computers in the inventory and 17 were sold, what is the value of the inventory left?

The inventory was at first 400*20

The cost of the computers sold is 400*17

Therefore, the expression we need to solve is 400*20-400*17

We can use the combining like terms formula so that x=400 (400 is the value that is the same in both parts of the expression so it should be x) getting:

 400*20-400*17 = 400(20-17) = 400*3 = 1,200

Solving with 2 formulas: the formula ax-bx=(a-b)x and the formula (a-b)x=ax-bx

In the following example we need to use the combining like terms formula ax±bx=(a±b)x and then use the distributing formula for removing parentheses (a±b)x=ax±bx.

The cost of one computer is 498 dollars. If there were 20 computers in the inventory and 5 were sold, what is the value of the inventory left?

The inventory was at first 498*20

The cost of the computers sold is 498*5

Therefore, the expression we need to solve is 498*20-498*5

We can use the combining like terms formula so that x=498 (498 is the value that is the same in both parts of the expression so it should be x) getting:

 498*20-498*5 = 498(20-5) = 498*15

To calculate the answer fast we can replace 498 with 500-2 getting (500-2)*15

Now we can use the distributing formula for removing parentheses: (a±b)x=ax±bx getting

(500-2)*15 = 500*15 – 2*15 = 500*10+500*5-30 = 5,000+2,500-30 = 7,500-30 = 7,420

 

Calculations with fractions formulas

Fractions addition and subtraction formula a⁄b±c⁄b=(a±c)⁄b

This formula involves adding or subtracting two fractions with a same denominator.

The formula has 2 forms (minus or plus):

a⁄b+c⁄b=(a+c)⁄b

a⁄b-c⁄b=(a-c)⁄b

Since the denominator is the same, we can add or subtract the nominators first and then divide the result by the denominator.

Consider the following example for using this formula with plus and minus signs:   

There pool was 13/15 filled with water. During the first week 3/15 of the water evaporated from the pool, during the second week 1/15 of the water evaporated from the pool. The poolman filled 6/15 of the pool with water. What part of the pool is filled with water?

We need to solve the expression 13/15-3/15-1/15+6/15, we need to use addition and subtraction formula: a⁄b±c⁄b=(a±c)⁄b getting

 13/15-3/15-1/15+4/15 = (13-3-1+6)/15 = 15/15 = 1 

The pull is full.

Fractions multiplication formula: a⁄b*c⁄d=a*c⁄b*d

Circle equations

Circle equations on the SAT test

SAT Subscore: Additional topics in math

Studying circle equations

On the SAT test circle equations topic is the last topic of additional topics in math that include 7 advanced topics (see the full topics list on the top menu). It is recommended to start learning additional topics in math with its first topic called complex numbers.

Circle equations topic is divided into sections from easy to difficult (the list of the sections appears on the left menu). Each section includes detailed explanations of the required material with examples followed by a variety of self-practice questions with solutions.

Finish studying heart of algebra subscore topics before you study this topic or any other additional topic in math. (Heart of algebra subscore includes basic algebra topics which knowledge is required for understanding additional topics in math). 

Circle equations- summary

A standard form of a circle contains 2 squares of binomials that are equal to a squared radius, so that a circle with a center (h,k) and a radius r has a standard form equation of (x-h)2+(y-k)2=r2.

To find the location of the circle in the xy plane, we need to draw it according to its standard equation.

If we open brackets of the standard form equation, we get the expended form: x2-2hx+y2-2ky+(h2+k2-r2)=0, in which there are 3 constants h2, k2 and -r2.

Rewriting the expended equation to a standard form: We can identify the first 4 components of the equation since each one of them has a different type of variable (x2,x,y2 and y) but the 3 constants will be given by a single number. We can find the 3 constants by completing 2 quadratic equations of x and y (finding k and h) with square of sum and square of difference formulas.

Continue reading this page for detailed explanations and examples.

A standard form equation of a circle in the xy plane

A standard form of a circle contains 2 squares of binomials that are equal to a squared radius, so that a circle with a center (h,k) and a radius r has a standard form equation of (x-h)2+(y-k)2=r2.

For example: If the center of the circle is located at the point (2,5) and the radius of the circle is equal to 3 then h=a and k=5 the standard equation will be (x-2)2+(y-5)2=32.

The figure below represents a circle in the xy plane and its standard equation.

The center of the circle O has the coordinates of (h,k) and its standard equation is (x-h)2+(y-k)2=r2.

A standard form equation of a circle

Drawing a circle in the xy plane using its standard form equation

We can be asked about the location of the circle, like in which quadrant the circle is located. To find the location of the circle in the xy plane, we need to draw it according to its standard equation.

Finding the location of the circle in the xy plane steps:
Step 1: Draw the axes and write the numbers of the quadrants in the xy plane.
Remember that the axes divide the coordinate plane into 4 quadrants, the first quadrant is the top right quadrant (for positive x values and positive y values) and moving counterclockwise. The quadrants are names using the Roman numbers I (+,+), II (-,+) , III (-,-) and IX(+,-).

Step 2: Draw the center of the circle (the center coordinates are h and k values from the equation).

Step 3: Draw the circle by adding the value of the radius around the center (the radius value is written in the equation).

Step 4: See in which quadrant the circle is located.

Consider the following example:

In which quadrant is located the circle that has an equation (x+5)2+(y+5)2=22?

The center of the circle coordinates are (-5,-5) and its radius is equal to 2.

The figure below presents the circle in the xy plane.

The circle is in the third quadrant.

Drawing a circle

An expanded equation of a circle

To learn more about quadratic equations on quadratic equations and quadratic functions page.

If we open brackets of the standard form equation, we get the expended form:
(x-h)2+(y-k)2=r2
x2+h2-2hx+y2+k2-2ky-r2=0

Note that there are 3 constants in the equations which are h2, k2 and -r2, therefore, we get the equation
x2-2hx+y2-2ky+(h2+k2-r2)=0

Finding a standard form equation of a circle from its expanded form

To find the center and the radius of the circle, we need to rewrite the expended equation to a standard form.

We can identify the first 4 components of the equation since each one of them has a different type of variable (x2,x,y2 and y) but the 3 constants will be given by a single number. We can find the 3 constants by completing 2 quadratic equations of x and y (finding k and h) with square of sum and square of difference formulas.

We know that the square of sum formula is (a+b)2=a2+b2+2ab.

We know that the square of difference formula is (a-b)2=a2+b2-2ab.

Consider the following example:

We are given the expression x2-6x.

What constant c should be added to this expression, so that we get a quadratic binomial in a form of (x-c)2?

We know that the square of difference formula is (a-b)2=a2+b2-2ab and we are given that a2=x2 and -2ab=-6x

a2=x2

x=a, therefore the first number is x

 

-2ab=-6x

We found that x=a therefore -2xb=-6x.

We can divide by -2x getting b=-6x/-2x.

b=3, therefore the second number is 3 and the quadratic binomial is (x-3)2

b2=9, therefore the constant that is added to the expression x2-6x to complete a quadratic binomial is 9.

Checking the answer:

(x-3)2=x2-6x+9

Consider the following example:

We are given the expression 2x2+20x,

What constant c should be added to this expression, so that we get a quadratic binomial in a form of (x-c)2?

First, we should factor out the number 2 to make the coefficient of x2 equal 1, so that the expression will be 2(x2+10x).

We know that the square of sum formula is (a+b)2=a2+b2+2ab and we are given that a2=x2 and 2ab=10x

a2=x2

a=x, therefore the first number is x

 

2ab=10x

We found that x=a, therefore 2bx=10x

b=5, therefore the second number is 5 and the quadratic binomial is 2(x+5)2

b2=25, therefore the constant that is added to the expression 2x2+20x to complete a quadratic binomial is 25*2=50.

Checking the answer:

2(x+5)2=2(x2+10x+25)

2(x+5)2=2x2+20x+50

Consider the following example:

What is the center and the radius of the circle equation x2-2x+y2-6y+1=0?

The circle standard equation is (x-h)2+(y-k)2=r2, where the center of the circle is (h,k) and the radius is r.

The given constant of 1 combines 3 constants so that c12+c22-c32=1
c1 for completing the expression x2-2x to quadratic binomial
c2 for completing the expression y2-6y to quadratic binomial
-c3 that is equal to the radius

Note that the sign of c3 is negative since the radius is transferred from the second side of the equation.

We need to complete x2-2x to x2-2x+c12=(x-c1)2 getting x2-2x+12=(x-1)2, therefore c1=1 and c12=1.

We need to complete y2-6y to y2-6y+c22=(x-c2)2 getting y2-6y+32=(y-3)2, therefore c2=3 and c22=9.

We know that c12+c22-c32=1, therefore 1+9-c32=1 and c32=9, c3=3.

x2-2x+1  +  y2-6y+9    +1-10  =0

  (x-1)2          (y-3)2          -9     =0

  (x-1)2          (y-3)2                  =32

The standard equation is (x-1)2+(y-3)2=32, therefore the center of the circle is (1,3) and the radius of the circle is 3.

Checking the answer:

(x-1)2+(y-3)2=32

x2+1-2x+y2+9-6y-9=0

x2-2x+y2-6y+1=0

Consider the following example:

What is the center and the radius of the circle equation 4x2-4x+9y2-12y+1=0?

4x2-4x+1  +  9y2-12y+9  +1-10  =0

   (2x-1)2           (3y-3)2        -9     =0

   (2x-1)2           (3y-3)2                =32

The standard equation is (2x-1)2+(3y-3)2=32, therefore the center of the circle is (1,3) and the radius of the circle is 3.

 

You just finished studying circle equations topic, the last topic of additional topics in math!

Angles, arc lengths and trig functions

Angles, arc lengths and trig functions on the SAT test

SAT Subscore: Additional topics in math

Studying angles, arc lengths and trig functions

On the SAT test Angles, arc lengths and trig functions topic is the sixth topic of additional topics in math that include 7 advanced topics (see the full topics list on the top menu). It is recommended to start learning additional topics in math with its first topic called complex numbers.

Before learning arc lengths and trig functions topic learn the topics circle theorems and right triangle trigonometry (from additional topics in math).

Arc lengths and trig functions topic is divided into sections from easy to difficult (the list of the sections appears on the left menu). Each section includes detailed explanations of the required material with examples followed by a variety of self-practice questions with solutions.

Finish studying heart of algebra subscore topics before you study this topic or any other additional topic in math. (Heart of algebra subscore includes basic algebra topics which knowledge is required for understanding additional topics in math). 

Angles, arc lengths and trig functions- summary

Angles, arc length and trig functions topic includes 3 parts:
Calculations of angles in radians.
Calculation of arc lengths and sector areas in radians.
Calculation of sine, cosine and tangent in radians.

A radian is defined as the angle subtended from the center of a circle which intercepts an arc equal in length to the radius of the circle. To find the number of radians in a circle, we need to divide the length of the circumference by the radius: 2πr/r=2π, meaning that the number of radians of arc in a circle is 2π.

The relationship between radian and degree measures:
2π radians is equal to 360 degrees, meaning that one radian is equal to 360/2π=180/π=180/3.14≈57 degrees.

Radian measure   =    degree measure
_________________        _________________
            π                                 180°

The relationship between central angle in radians, arc length and sector area:
central angle    =         arc length              =     sector area
_____________       ____________________         ____________
        2π                   circle circumference           circle area

Special right triangles in circles:
In these questions we are given a circle which center is located at the axis intersection point (0,0).

Special right triangles are right triangles whose sides are in a particular ratio.
Two special right triangles are 30°, 60°, 90° triangle and 45°, 45°, 90° triangle.

The radian measures of angles of special right triangles are:
30° angle radian measure π/6; 45° angle radian measure π/4, 60° angle radian measure π/3; 90° angle radian measure: π/2.

We can draw a right triangle from any point on the circle so that the hypotenuse is equal to the radius of the circle and the sides of the triangle are equal to x and y coordinates of the point.

We can check the ratios between the sides of the triangle to see if the triangles are special right triangles. If the triangles are special right triangles, we know the values of their angles.

Calculating trigonometric functions (sin, cos and tan) with radian angle measures in right triangles in circles:
We calculate trigonometric functions under the assumption of unit circle, meaning that the radius is equal to 1. Since the hypotenuse is equal to the radius, we know that the hypotenuse is equal to 1.

The trigonometric functions values are:
sin(A)=opposite/hypotenuse= opposite/1=opposite.
cos(A)=adjacent/hypotenuse= adjacent/1=adjacent.
tan(A)=opposite/adjacent.

Continue reading this page for detailed explanations and examples.

The relationship between radian and degree measures

Degrees in a circle: The number of degrees of arc in a circle is 360.

Radians in a circle: To find the number of radians in a circle, we need to divide the length of the circumference by the radius: 2πr/r=2π, meaning that the number of radians of arc in a circle is 2π.

Note that the underlined sentences above are provided at the beginning of each SAT math section.

Therefore, the connection is that 2π radians is equal to 360 degrees, meaning that one radian is equal to 360/2π=180/π=180/3.14≈57 degrees.

The figure below presents the connection between radian and degree measures.

A radian is defined as the angle subtended from the center of a circle (marked in red) which intercepts an arc equal in length to the radius of the circle (the radii and the equal arc are marked in blue).

The connection between radians and degrees

Since 2π radians is equal to 360 degrees, we can calculate radian measure given degree measure or calculate degree measure given radian measure using the following ratios:

Radian measure   =    degree measure
________________        _________________
           2π                               360°

We can simplify the proportion getting:

Radian measure   =    degree measure
_________________        _________________
             π                                 180°

Consider the following example:

Covert 100° to radians.

Radian measure   =    degree measure
________________         ________________
             π                                 180°

Represent radian measure by the variable x.

x    =    100
__         ____
π          180

180x=100π
x=100/180π
x=0.55π=0.55*3.14=1.74 radians
100 degrees are equal to 0.55π radians.

Consider the following example:

Covert 3.5 radians to degrees.

Radian measure   =    degree measure
________________         ________________
             π                                 180°

Represent degree measure by the variable x.

3.5   =    x
___        ___
 π          180

πx=180*3.5
x=180*3.5/π
x=200°
3.5 radians are equal to 200 degrees.

We can also calculate without using the proportion using the fact that 1 radian is equal to 57 degrees:
3.5*57=200

Calculating angles, arc length and sector areas with radians

We can measure arc length and sector areas with radians instead of degrees.

We know that the relationship between central angle in degrees, arc length and sector area is given by the following ratios:

central angle    =            arc length              =     sector area
_____________        _____________________          ____________
       360°                  circle circumference            circle area

We also know that the number of radians of arc in a circle is 2π, therefore we can substitute 360 degrees by 2π:

The relationship between central angle in radians, arc length and sector area is given by the following ratios:

central angle    =         arc length              =     sector area
_____________       ____________________         ____________
          2π                circle circumference           circle area

Consider the following example:

The central angle of a circle is equal to 0.5π, the circumference of the circle is equal to 10 centimeters.

What is the measure of the arc formed by this angle?

What is the measure of the sector area formed by this angle?

Calculating the arc length:

central angle    =         arc length
_____________        ______________________
         2π                    circle circumference

Represent arc length by the variable x and plug the given data into the ratios equation:

0.5 π  =   x
____     ___
2π        10

x=0.5*10/2
x=2.5
The arc length is 2.5 centimeters.

Checking the answer:
0.5 π  = 2.5
____     ___
2π        10

0.5/2=2.5/10
1/4=1/4

Calculating the sector area:
The circle area formula is A=πr2.
A=π*102
A=100π
A=100*3.14=314

central angle    =    sector area
_____________        __________
2π                      circle area

Represent sector area by the variable x and plug the given data into the ratios equation:

0.5π  =     x
____      _____
2π         100π

x=0.5*100π/2
x=25π
The sector area is 25π.

Checking the answer:
0.5π  =   25π
____      _____
2π         100π

0.5/2=25/100
1/4=1/4

Special right triangles in circles

In these questions we are given a circle which center is located at the axis intersection point (0,0)

We can draw a right triangle from any point on the circle so that the hypotenuse is equal to the radius of the circle and the sides of the triangle are equal to x and y coordinates of the point.

We can check the ratios between the sides of the triangle to see if the triangles are special right triangles. If the triangles are special right triangles, we know the values of their angles. 

Special right triangles measures

Special right triangles are right triangles whose sides are in a particular ratio.
Two special right triangles are 30°, 60°, 90° triangle and 45°, 45°, 90° triangle.
Special right triangles with their side sizes length are given at the beginning of each SAT section.

30°, 60°, 90° triangle:
In a 30°, 60°, 90° right triangle the side opposite the 30° angle is half the length of the hypotenuse and the side opposite to 60° angle is equal to the length of a side opposite to 30° angle multiplied by √3.
In 30°, 60°, 90° triangle the sides are x, x√3 and 2x.

 45°, 45°, 90° triangle:
In a 45°, 45°, 90° right triangle the sides opposite the 45° angles are equal and the hypotenuse is equal to the side opposite to 45° angle multiplied by √2.
In 45°, 45°, 90° triangle the sides are s, s and s√2.

The following graphs present the special right triangles with the side sizes length.

Two special right triangles are 30°, 60°, 90° triangle and 45°, 45°, 90° triangle.

Radian measures of angles of special right triangles

30° angle radian measure: π/6

Since 2π radians is equal to 360 degrees we get:

Radian measure   =    degree measure
________________         ________________
             π                                180°

x   =  30
__     ___
π      180

x=30π/180
x=π/6

45° angle radian measure: π/4

Since 2π radians is equal to 360 degrees we get:

Radian measure   =    degree measure
________________         ________________
             π                                 180°

Representing radian measure by the variable x and plugging the degree measure into the proportion:

x  = 45
__   ___
π    180

x=45π/180
x=π/4

Since 30*1.5=45 we can also multiply the radian measure of 30° angle by 1.5 getting:
1.5*π/6= π/4.

60° angle radian measure: π/3

Since 2π radians is equal to 360 degrees we get:

Radian measure   =    degree measure
________________         ________________
             π                                 180°

Representing radian measure by the variable x and plugging the degree measure into the proportion:

x  = 60
__   ___
π    180

x=60π/180
x=π/3

Since 30*2=60 we can also multiply the radian measure of 30° angle by 2 getting:
2*π/6= π/3.

90° angle radian measure: π/2

Since 2π radians is equal to 360 degrees we get:

Radian measure   =    degree measure
________________         ________________
              π                               180°

Representing radian measure by the variable x and plugging the degree measure into the proportion:

x  = 90
__   ___
π    180

x=90π/180
x=π/2

Since 30*3=90 we can also multiply the radian measure of 30° angle by 3 getting:
3*π/6= π/2.

The following figures present special right triangles with their side lengths and angles in degrees and radians (the radian measures are marked in red).

Special right triangles radian and degree values

Calculating side lengths and radian angle measures in special right triangles in circles

We are given a circle which center is located at the axis intersection point (0,0)

We can draw a right triangle from any point on the circle so that the hypotenuse is equal to the radius of the circle and the sides of the triangle are equal to x and y coordinates of the point.

We can check the ratios between the sides of the triangle to see if the triangles are special right triangles. If the triangles are special right triangles, we know the values of their angles. 

Consider the following example:

The points coordinates are A(3,3) and b(-4,4/√3). Both points are located on a circle and the center of the circle is located at the axis intersection point (0,0).

What is the size of the angle BOD?

What is the size of the angle AOC?

Finding side lengths and radian angle measures in special right triangles in circles

The figure above presents 2 points A and B that are located on a circle. The center of the circle is located at the axis intersection point (0,0). The coordinates of point A are (3,3) and the coordinates of point B are (-4,4/√3).

Drawing right triangles from points on a circle and calculating side length from the coordinates of the points:

The lines AO and BO are radii of the circle. 

The line AC in drawn from point A to create a right triangle ACO, so that the angle ACO is equal to 90°. Since the angle ACO is equal to 90°, the length of the side AC is equal to y coordinate of point A so that AC=3. In addition, the length of the side CO is equal to x coordinate of point A so that CO=3.

The line BD in drawn from point A to create a right triangle BDO, so that the angle BDO is equal to 90°. Since the angle BDO is equal to 90°, the length of the side BD is equal to the y coordinate of point B so that BD=4/√3. In addition, the length of the side DO is equal to the absolute value of the x coordinate of point B so that DO=|-4|=4.

Calculating the angles of special right triangles:

In the triangle ACO the side lengths are AO=CO=3, therefore the triangle ACO is an isosceles triangle. An isosceles triangle is a special right triangle and we know its angles measures are 45°,45° and 90° and the radian measures are π/4, π/4 and π/2.

Note that we can also calculate the angle measures: Since the triangle ACO is an isosceles triangle, the angles CAO and AOC are equal. Since the angle ACO is equal to 90° and the sum of the angles in a triangle is 180°, the angles CAO and AOC are equal to 45°.

In the triangle BDO the side lengths are BD=4/√3 and DO=4, therefore the triangle ACO is a special right triangle and its angles measures are 30°,60° and 90° and the radian measures are π/6, π/3 and π/2. Therefore, the angle BOD=30°= π/6 and the angle DBO=60°=π/3.

Calculating trigonometric functions (sin, cos and tan) with radian angle measures in right triangles in circles

Special right triangles with their side sizes length are given at the beginning of each SAT section.

The following figure presents 2 special right triangles and their angles in degrees, like given in the SAT (note that the radian measures are not given).

Two special right triangles are 30°, 60°, 90° triangle and 45°, 45°, 90° triangle.

We calculate trigonometric functions under the assumption of unit circle, meaning that the radius is equal to 1. Since the hypotenuse is equal to the radius, we know that the hypotenuse is equal to 1.

Remember the trigonometric functions values:
sin(A)=opposite/hypotenuse= opposite/1=opposite.
cos(A)=adjacent/hypotenuse= adjacent/1=adjacent.
tan(A)=opposite/adjacent.

To learn more about trigonometric functions go to right triangle trigonometry page.

Calculating the radian angles measures of triangles

At the beginning of each SAT math section, it is given that:
The number of degrees of arc in a circle is 360.
The number of radians of arc in a circle is 2π.

Therefore, we know that 360°=2π and π=180°.
The angle of 30°: 30=360/12=2π/12=π/6.
The angle of 45°: 45=360/8=2π/8=π/4.
The angle of 60°: 60=360/6=2π/6=π/3.
The angle of 90°: 90=360/4=2π/4=π/2.
The angle of 120°: 120=360/3=2π/3.
The angle of 135°: 135=360*3/8=2π*3/8=3π/4.
The angle of 180°: 180=360/2=2π/2=π.

Special right triangle with 30°, 60°, 90° angles

Calculating the side lengths of the triangle:
We know the side lengths from the beginning of each math SAT section: x, 2x and x√3. Since the hypotenuse is equal to 1 (unit circle) we know that 2x=1 and x=1/2. Therefore, the sides are 1/2, √3/2 and 1.

Calculating the sine, cosine and tangent of the angle of π/6 radians:
We need to look at the given 30°, 60°, 90° special right triangle, we saw that its sides are 1/2, √3/2 and 1.
sin (π/6)=opposite/1=(1/2)/1=1/2.
cos (π/6)=adjacent/1=(√3/2)/1=√3/2.
tan (π/6)=opposite/adjacent=(1/2)/(√3/2)=1/√3 multiply by √3/√3 getting (1*√3)/(√3*√3)= √3/√9=√3/3.

Calculating the sine, cosine and tangent of the angle of π/3 radians:
We need to look at the given 30°, 60°, 90° special right triangle, we saw that its sides are 1/2, √3/2 and 1.
sin(π/3)=opposite/1=√3/2.
cos(π/3)=adjacent/1=1/2.
tan(π/3)=opposite/adjacent=(√3/2)/(1/2)= √3.

Special right triangle with 45°, 45°, 90° angles

Calculating the side lengths of the triangle:

We know the side lengths from the beginning of each math SAT section: s, s and s√2. Since the hypotenuse is equal to 1 (unit circle) we know that s√2=1 and s=1/√2. Therefore, the sides are 1/2, 1/2 and 1.

Calculating the sine, cosine and tangent of the angle of π/4 radians:
We need to look at the given 45°, 45°, 90° special right triangle.
sin(π/4)=opposite/1=1/√2 multiplying by √2/√2 getting (1*√2)/( √2*√2)=√2/2.
cos(π/4)=adjacent/1=1/√2 multiplying by √2/√2 getting (1*√2)/( √2*√2)=√2/2.
tan(π/4)=opposite/adjacent=(√2/2)/(√2/2)=1.

Angle measures of 0 radians and π/2 radians

Calculating the sine, cosine and tangent of the angle of 0 radians:
sin(0)=opposite/1: If the angle is close to 0, the side that is opposite to the angle is also close to 0, therefore sin(0)=0/1=0.
cos(0)=adjacent/1= If the angle is close to 0, the side that is adjacent to the angle is almost equal to the hypotenuse which is equal to 1, therefore cos(0)=1/1=1.
tan(0)=opposite/adjacent= If the angle is close to 0, the side that is opposite to the angle is also close to 0 and the side that is adjacent to the angle is equal to the hypotenuse which is equal to 1 therefore tan(0)=0/1=0.

Calculating the sine, cosine and tangent of the angle of π/2 radians:
We know that 2π=360°, therefore π/2=90°.
sin(π/2)=opposite/1=If the angle is close to 90°, the side that is opposite to the angle is almost equal to the hypotenuse which is equal to 1, therefore sin(π/2)=1/1=1.
cos(π/2)=adjacent/1= If the angle is close to 90°, the side that is adjacent to the angle is close to 0, therefore cos(π/2)=0/1=0.
tan(π/2)=opposite/adjacent= If the angle is close to 90°, the side that is opposite to the angle is 1 and the side that is adjacent to the angle is 0. We can’t divide by 0, therefore tan(π/2) is not defined.

Angle measures bigger than 90°:

We can convert these angles to angles smaller than 90° using 2 formulas:
sin(α)=sin(180-α) or with radians: sin(α)=sin(π-α).
cos(α)=-cos(180-α) or with radians: cos(α)=-cos(π-α).
We also know that tan(α)=sin(α)/cos(α)

Calculating the sine, cosine and tangent of the angle of 2π/3 radians:
We know that 2π=360°, therefore 2π/3=360°/3=120°>90°.
sin(2π/3)= sin(π-2π/3)=sin{(3π-2π)/3}=sin(π/3), we found that sin(π/3)=√3/2.
cos(2π/3)=-cos(π-2π/3)= -cos{(3π-2π)/3}=-cos(π/3), we found that cos(π/3)=1/2, therefore -cos(π/3)=-1/2.
tan(2π/3)= sin(2π/3)/cos(2π/3)=(√3/2)/(-1/2)=-(√3/2)*2=-√3.

Calculating the sine, cosine and tangent of the angle of 3π/4 radians:
We know that 2π=360°, therefore π=180° and 3π/4=180°*3/4=135°>90°.
sin(3π/4)= sin(π-3π/4)=sin{(4π-3π)/4}=sin(π/4), we found that sin(π/4)=√2/2.
cos(3π/4)=-cos(π-3π/4)=-cos{(4π-3π)/4}=-cos(π/4), we found thar cos(π/4)=√2/2, therefore -cos(π/4)=-√2/2.
tan(3π/4)=sin(3π/4)/cos(3π/4)=(√2/2)/(-√2/2)=-1.

Calculating the sine, cosine and tangent of the angle of π radians:
Since 2π=360°, the angle of π radians is equal to 180°.
sin(π)=sin(π-π)=sin(0)=0.
cos(π)=-cos(π-π)=-cos(0)=-1.
tan(π)=sin(π)/cos(π)=0/-1=0.

You just finished studying angles, arc lengths and trig functions topic, the sixth topic of additional topics in math!

Continue studying the next additional topic in math- circle equations.

Circle theorems

Circle theorems on the SAT test

SAT Subscore: Additional topics in math

Studying circle theorems

On the SAT test circle theorems topic is the fifth topic of additional topics in math that include 7 advanced topics (see the full topics list on the top menu). It is recommended to start learning additional topics in math with its first topic called complex numbers.

Circle theorems topic is divided into sections from easy to difficult (the list of the sections appears on the left menu). Each section includes detailed explanations of the required material with examples followed by a variety of self-practice questions with solutions.

Finish studying heart of algebra subscore topics before you study this topic or any other additional topic in math. (Heart of algebra subscore includes basic algebra topics which knowledge is required for understanding additional topics in math). 

Circle theorems- summary

Circle theorems topic includes two parts:
Calculating arc length and sector areas using central angles measured in degrees.
Calculating angle measures in degrees inside a circle.

The following formulas are provided at the beginning of each SAT math section:
Circumference of a circle formula is C=2πr.
Area of a circle formula is A=πr2.
Number of degrees of arc in a circle is 360.

Pi (π) is the ratio of the circumference of any circle to the diameter of that circle. Regardless of the circle’s size, this ratio will always equal π (approximately 3.14).

A triangle that one of its angles is a central angle has 2 radii as its sides and it is therefore an isosceles triangle (one of its angles is a central angle and the other 2 angles are equal).

The intersection point of the diameters is at the center of the circle and it divides each diameter to 2 radii. The intersection creates 4 central angles: 2 pairs of equal vertical angles and 4 pairs of supplementary angles (the sum of supplementary angles is 180°).

The relationship between central angle, arc length and sector area is given by the following ratios:

central angle    =             arc length              =     sector area
______________       _____________________          ____________
         360°                 circle circumference            circle area

Continue reading this page for detailed explanations and examples.

Calculating circumference and area of a circle

The following formulas are provided at the beginning of each SAT math section:
Circumference of a circle formula is C=2πr.
Area of a circle formula is A=πr2.

Units of measurement:
Arc length and circumference measure distance in units such as inches or centimeters.
Area and sector measure area is square units such as square inches or square centimeters.
Angles are measured in degrees.

Consider the following example:

The radius of the circle is equal to 10 centimeters.

What are the circumference and the area of the circle?

Circumference of a circle:
The circumference formula is C=2πr.
C=2*10*π=20π

Area of a circle:
The area formula is A=πr2.
A=πr2=π*102=100π

The relationship between central angle, arc length and sector area

A central angle has a vertex at the center of a circle O and sides located on the circle circumference in two points A and B. The central angle ∠AOB determines a portion of the circumference (an arc) AC and a potion from the area (sector).

A minor arc: Each central angle divides a circle into two arcs. The smaller of the two arcs is called the minor arc and the larger of the two arcs is called the major arc (a minor arc subtends an angle less than or equal to 180°). Note that when dealing with arcs, we always look at the minor arc.

The relationship between central angle, arc length and sector area is given by the following ratios:

central angle    =             arc length              =     sector area
______________       _____________________          ____________
         360°                 circle circumference            circle area

This relationship presents 3 equal ratios:

Ratio 1: The angle as a portion of the total degrees of the circle (360°) =

Ratio 2: The arc length as a portion from the length of the circle circumference (C=2πr) =

Ratio 3: The sector area as a portion from the total area of the circle (A=πr2).

Note that we can calculate the denominators of the arc and the sector ratios given the value of the radius r.

In the figure below we have the following ratios:

   ∠AOC     =        arc AC          =    sector ACO area
   _______      _______________        ________________
     360°         circumference               circle area

 

   ∠AOD     =          arc AD           =    sector ADO area
   ________     ________________        _________________
     360°            circumference             circle area

 

   ∠BOD     =         arc BD            =     sector BDO area
  ________      _______________          __________________
     360°           circumference                circle area

 

   ∠BOC     =          arc BC           =     sector BCO area
  ________       ______________           _________________
     360°          circumference                 circle area

The ratios formula: central angle, arc length and sector area

Consider the following example:

A central angle of a circle is equal to 120 degrees and the radius of the circle is equal to 5 centimeters.

What is the arc length of the central angle?

Calculating the circumference of the circle:
C=2πr
C=2*5*π=10π

Calculating the arc length with the ratios:
Ratio 1: The angle as a portion of the total degrees of the circle (360°).
Ratio 2: The arc length as a portion from the length of the circle circumference (C=2πr).

central angle     =           arc length
______________        _____________________
        360°                  circle circumference

Representing with the variable x the arc length and plugging the given values into the ratio:

120°   =     x
_____      _____
360°        10π

360x=1200π
x=1200/360π
x=120/36π
x=10/3π=3.33π
x=31/3π=31/3*3.14=10.47
The arc length is 10.47 centimeters.

Checking the answer:
120/360=1/3=0.33
3.33/10.47=0.32
The answers are different because of rounding differences.

Consider the following example:

The arc length of a circle is 2 centimeters and the radius of the circle is 1.6 centimeters.

What is the area of the sector of the arc?

Calculating the circumference of the circle:
C=2πr
C=2*1.6*π=3.2π=10.

Calculating the area of the circle:
The area formula is A=πr2.
A=π*1.62=2.56π=8.

Calculating the sector area with the ratios:
Ratio 1: The arc length as a portion from the length of the circle circumference (C=2πr).
Ratio 2: The sector area as a portion from the total area of the circle (A=πr2).

Representing with the variable x the sector area and plugging the given values into the ratio:

       arc length              =     sector area
____________________        _____________
circle circumference           circle area

 1.6   =   x
____      ___
 10         8

1.6*8=10x
x=1.6*8/10
x=1.28
The sector area is 1.28 centimeters.

Checking the answer:
1.6/10=0.16
1.28/8=0.16

Angle relationships in circle

This subject includes 2 angle types: 

Angles of isosceles triangles in a circle.

Angles created by intersection of diameters in a circle.

Angles of isosceles triangles in a circle

At the beginning of each SAT math section, it is provided that the number of degrees of arc in a circle is 360. In other words, the sum of central angle measures in a circle is equal to 360°.

A triangle that one of its angles is a central angle has 2 radii as its sides and it is therefore an isosceles triangle. Meaning that one of its angles is a central angle and the other 2 angles are equal.

The sum of angles in a triangle is 180°, if we are given the value of the central angle or the value of one of the equal angles, we can calculate the other angle values.

Consider the following example:

In the figure below, the point O is the center of a circle. The angle ∠AOB is equal to 100°.

What is the value of the angle ∠OAB?

Angle relationships in circle-Isosceles triangles- example

Since the angle ∠AOB is a central angle, the sides OA and OB are radii of the circle and therefore the triangle ABO is an isosceles triangle.

In an isosceles triangle the angles opposite the two equal sides are equal, therefore ∠OAB=∠ABO.

The sum of angles in a triangle is equal to 180°, therefore ∠OAB+∠ABO+∠AOB=180°.

We are given that ∠AOB=100° and we know that ∠OAB=∠ABO, therefore 100°+∠OAB+∠ABO=180°.

Represent by x the equal angles ∠OAB and ∠ABO we get an equation that we can solve:
100+2x =180
2x=80
x=40
The value of the angle ∠OAB is 40 degrees.

Checking the answer:
40+40+100=180
180=180

Consider the following example:

In the figure below, the point O is the center of a circle.
The angle ∠AOB is twice bigger than the angle ∠COA.
The angle ∠COB is 5 times bigger than the angle ∠COA (the angle ∠COB is bigger than 180°).

What is the value of the angles ∠AOB and ∠COB?

Angle relationships in circle-central angle in a circle

Represent by x the angle ∠COA.

We are given that the angle ∠AOB is twice bigger than the angle ∠COA, therefore the angle ∠AOB=2x.
We are given that the angle ∠COB is 5 times bigger than the angle ∠COA, therefore the angle ∠COB=5x.

The sum of central angle measures in a circle is equal to 360° and we are given that the point O is the center of the circle, therefore we can write an equation and solve it:
x+2x+5x=360
8x=360
x=45

We know that the angle ∠AOB=2x, therefore the angle ∠AOB=90°.
We know that the angle ∠COB=5x, therefore the angle ∠COB=5*45=225 degrees.

Angles created by intersection of diameters in a circle

The intersection point of the diameters is at the center of the circle and it divides each diameter to 2 radii.

The intersection creates 4 central angles: 2 pairs of equal vertical angles and 4 pairs of supplementary angles (the sum of supplementary angles is 180°).

Given the value of one of the 4 central angles, we can calculate the other central angles.

In the figure below AB and CD are diameters of the circle.

What are the central and supplementary angles created by the diameters?

Angles created by intersection of diameters

The intersection creates 4 equal radii AO, CO, DO and BO.

The intersection point O is at the center of the circle and it divides each diameter to 2 radii: the diameter AB is divided to radii AO and OB; the diameter CD is divided to radii CO and OD.

The intersection creates 4 central angles: ∠COA, ∠AOD, ∠DOB and ∠BOC.

The intersection creates 2 pairs of equal vertical angles: ∠COA=∠BOD; ∠COB=∠AOD.

The intersection creates 4 pairs of supplementary angles: ∠COA+∠AOD=180°; ∠AOD+∠DOB=180°;  ∠DOB+∠BOC=180°; ∠BOC+∠COA=180°.

 Consider the following example:

In the figure below, O the is center of the circle and the chords AB and CD intersect at point O. The angle ∠BOD is equal to 40°.

What is the ratio between the angles ∠ACO and ∠ADO?

Example- angles created by intersection of diameters in a circle

Given the value of one of the 4 central angles, we can calculate the other central angles:

We are given that ∠BOD=40°

∠AOC=∠BOD (vertical angles are equal), therefore, ∠AOC=40°.

∠BOD+∠AOD=180° (the sum of supplementary angles is 180°), therefore ∠AOD=180°-40°=140°.

Calculating the angles in triangle ACO:

We are given that O the is center of the circle and the chords AB and CD intersect at point O, therefore AB and CD are diameters and create 4 radii CO=BO=AO=DO.

Since CO=AO the triangle ACO is isosceles and the angles ∠ACO and ∠CAO are equal (represented by x).

The sum of angles in a triangle is equal to 180°, therefore ACO+CAO+AOC=180°.

Plugging the values we get an equation that we can solve:
x+x+40=180
2x=140
x=70
The angle ∠ACO is equal to 70 degrees.

Calculating the angles in triangle ADO:

We are given that O the is center of the circle and the chords AB and CD intersect at point O, therefore AB and CD are diameters and create 4 radii CO=BO=AO=DO.

Since AO=DO the triangle ADO is isosceles and the angles ∠DAO and ∠ADO are equal (represented by x).

The sum of angles in a triangle is equal to 180°, therefore DAO+ADO+AOD=180°.

We found that the angle ∠ACO is equal to 140 degrees.

Plugging the values, we get an equation that we can solve:
x+x+140=180
2x=40
x=20
The angle ∠ADO is equal to 20 degrees.

The ratio between the angles ∠ACO and ∠ADO is 70/20=7/2=31/2

Note that we need to write the value of an angle ∠ACO in the nominator and not the opposite.

You just finished studying circle theorems topic, the fifth topic of additional topics in math!

Continue studying the next additional topic in math- angles, arc lengths and trig functions.

Complex numbers

Complex numbers on the SAT test

SAT Subscore: Additional topics in math

Studying complex numbers

On the SAT test complex numbers topic is the first topic of additional topics in math that include 7 advanced topics (see the full topics list on the top menu). 

Complex numbers topic is divided into sections from easy to difficult (the list of the sections appears on the left menu). Each section includes detailed explanations of the required material with examples followed by a variety of self-practice questions with solutions.

Finish studying heart of algebra subscore topics before you study this topic or any other additional topic in math. (Heart of algebra subscore includes basic algebra topics which knowledge is required for understanding additional topics in math). 

Complex numbers- summary

A complex number has a real component and an imaginary component, it is written in a form of a + bi, where a and b are real numbers, and i is an imaginary number satisfying i2 = −1.

An imaginary number is a number that, when squared, has a negative result (i2 = −1). Since an imaginary number i is equal to a square root of a negative number -1 it does not have a tangible value (negative numbers don’t have real square roots since a square is either positive or zero).

The unit imaginary number, i, equals the square root of minus 1, so that i=√-1. As said above, when squared imaginary number has a negative result, so that i2=(√-1)2=-1.

For example: 3i is an imaginary number, and its square is (3√-1)2=9*-1=-9.

To add or subtract complex numbers, combine like terms (real terms with real terms and imaginary terms with imaginary terms). 

To multiply complex numbers, multiply the numbers with foil formula, replace i2 with -1 and combine like terms.

To divide complex numbers, you need to cancel the denominator by turning the imaginary component in the denominator to a real number. This is done by multiplying the numerator and the denominator by the conjugate of the denominator. The next steps are multiplying the numbers with foil formula, replacing i2 with -1 and combining like terms.

Continue reading this page for detailed explanations and examples.

Adding and subtracting complex numbers

Combine like terms: real terms with real terms and imaginary terms with imaginary terms and write the result as a+bi. 

Consider the following example:

If a=6+4i and b=2i-4, that are the values of a-b and a+b?

a=6+4i b=2i-4 a-b=6+4i-(2i-4)= 6+4i-2i+4=10+2i a+b=6+4i+(2i-4)= 6+4i+2i-4=2+6i

Multiplying complex numbers

Remember that since i=√-1, the value of i2 is i2=-1.

If after multiplying we get i2, we can write it as -1 and continue solving.

Steps for multiplying complex numbers:

Step 1: Multiply the numbers with foil formula.
The FOIL formula is y=(x+c)(x+d)= x2+dx+cx+cd= x2+(c+d)x+cd.

Step 2: Replace i2 with -1.

Step 3: Combine like terms (real terms with real terms and imaginary terms with imaginary terms) and write the result as a+bi.

Consider the following example:

If a=6+4i and b=2i-4, that is the value of a*b?

Step 1: Multiplying the numbers with foil formula:

a=6+4i

b=2i-4

a*b=(6+4i)(2i-4)=12i-24+8i2-16i

Step 2: Replacing i2 with -1:

a*b=12i-24+8i2-16i

a*b=12i-24+8*-1-16i

Step 3: Combining like terms:

a*b=-32-4i

Dividing complex numbers

We have a numerator and a denominator as 2 complex numbers in a form of a+bi and we need to simplify the result to a form of one complex number in a form of a+bi (staying without the denominator).

To cancel the denominator, we need to turn the imaginary component in the denominator to a real number, this is done by multiplying the numerator and the denominator by the conjugate of the denominator.

 For example:
We learned in the quadratic equations topic that (a+b)(a-b)=a2-b2.
If we multiply the complex number a+bi by a conjugate of a-bi we get (a+bi)(a-bi)=a2-b2i2.
Since we know that i2=-1. The expression a2-b2i2 is equal to a2+b2. This outcome is a real number.

Steps for dividing complex numbers:

Step 1: Multiply the numerator and the denominator by the conjugate of the denominator (conjugate divided by itself is equal to 1 and we can multiply by 1 without changing the original value).

Step 2: Multiply the numbers in the numerator and the denominator with foil formula.

Step 3: Replace i2 in the numerator and the denominator with -1. In the denominator you will be left with real terms without imaginary terms.

Step 4: Combine like terms and write the answer as a complex number in the numerator in a form of a+bi divided by a real number in the denominator.

Consider the following example:

If a=6+4i and b=2i-4, that is the value of a/b?

Step 1: Multiplying the numerator and the denominator by the conjugate of the denominator:

a=6+4i

b=2i-4

a   = 6+4i
__    _____
b      2i-4

The conjugate of the denominator is 2i+4.

a    =  6+4i   =  (6+4i)(2i+4)
__      _____       ____________
b         2i-4        (2i-4) (2i+4)

Step 2: Multiplying the numbers in the numerator and the denominator with foil formula:

a   =  12i+24+8i2+16i
__      _______________
b              4i2-16

Step 3: Replacing i2 in the numerator and the denominator with -1:

a   =  12i+24+8i2+16i  =   12i+24+8*-1+16i
__     ________________       ________________
b               4i2-16                        4*-1-16

Step 4: Combining like terms and writing the answer as a complex number in the numerator divided by a real number in the denominator:

a      =    12i+24+8*-1+16i   =   28i+16   =    28i  +  16  =   -7i  –  4
__           _________________       _______       ____     ____     ___    ___
b                    4*-1-16                     -20            -20      -20        5       5

You just finished studying complex numbers topic, the first topic of additional topics in math!

Continue studying the next additional topic in math- volume word problems.

Congruence and similarity

Congruence and similarity on the SAT test

SAT Subscore: Additional topics in math

Studying congruence and similarity

On the SAT test congruence and similarity topic is the third topic of additional topics in math that include 7 advanced topics (see the full topics list on the top menu). It is recommended to start learning additional topics in math with its first topic called complex numbers.

Congruence and similarity topic is divided into sections from easy to difficult (the list of the sections appears on the left menu). Each section includes detailed explanations of the required material with examples followed by a variety of self-practice questions with solutions.

Finish studying heart of algebra subscore topics before you study this topic or any other additional topic in math. (Heart of algebra subscore includes basic algebra topics which knowledge is required for understanding additional topics in math). 

Congruence and similarity- summary

Congruence and similarity questions include congruent angles and similar triangles topics.

Similar triangles have the same angle measures and their corresponding side lengths are related by a constant ratio.

Sum of angles in a triangle: The sum of the measures in degrees of the angles of a triangle is 180.

Angle measures in triangles: In an isosceles triangle the angles opposite the two equal sides are equal; the angles of an equilateral triangle are equal to 60°.

Congruent angles are angles that have the same angle measure.

Vertical Angles are the angles opposite each other when two lines cross, vertical angles are equal.

Supplementary angles are those angles that measure up to 180 degrees. Angles that lie on the same side of a straight line, are always supplementary.

Alternate angles are angles located in opposite positions relative to a transversal intersecting two parallel lines. Alternate angles are equal.

Corresponding angles are angles located in the same position relative to parallel lines intersected by a transversal. Corresponding angles are equal.

Two intersecting lines create vertical angles and supplementary angles.

Two parallel lines combined with two intersecting lines form 2 similar triangles and equal alternate angles.

A parallel line inside a triangle forms 2 similar triangles and equal corresponding angles.

Continue reading this page for detailed explanations and examples.

Angle relationships inside a triangle

Angle relationships inside a triangle

Sum of angles in a triangle:
The sum of the measures in degrees of the angles of a triangle is 180. (This statement is provided at the beginning of each SAT math section).
If use the variables x°, y° and z° to represent angles then x°+y°+z°=180°.

Angle measures in an isosceles triangle:
An isosceles triangle is a triangle that has two equal sides, the angles opposite the two equal sides are equal.

Angle measures in an equilateral triangle:
An equilateral triangle is a triangle with all three sides of equal length, the angles of an equilateral triangle are equal to 60°.

The figure below presents a triangle and its angles.

Angle sum inside a triangle

Consider the following example:

In a triangle ABC given that the angle A is equal to 50° and AB=BC.

What is the value of the angles B and C?

The sum of the measures of the angles of a triangle is 180°, therefore the angles A+B+C=180°.
We are given that AB=BC, therefore the angle B is equal to the angle C.
We are given that A=50°.

Therefore, we can represent by x the value of angle B and C and solve an equation with 1 variable:
x+x+50°=180°
2x=180°-50°
2x=130°
x=65°

Checking the answer:
65°+65°+50°=180°
180°=180°

Consider the following example:

In a triangle ABC given that the angle A is equal to 60° and the angle B is twice bigger than the angle C.

What is the value of the angles B and C?

The sum of the measures of the angles of a triangle is 180°, therefore the angles A+B+C=180°.
We are given that A=60°.
We are given that the angle B is twice bigger than the angle C.

Therefore, we can represent by x the value of angle C and solve an equation with 1 variable:
2x+x+60°=180°
3x=120°
x=40° therefore C=40° and D=40°*2=80°

Checking the answer:
40°+80°+60°=180°
180°=180°

Angle relationships between intersecting lines- vertical and supplementary angles

Two intersecting lines create vertical angles and supplementary angles.

Vertical Angles are the angles opposite each other when two lines cross (vertical means that they share the same vertex). Vertical angles are equal.

Supplementary angles are those angles that measure up to 180 degrees. Angles that lie on the same side of a straight line, are always supplementary.

The figure below shows the angles created between two intersecting lines.

Angle relationships- vertical and supplementary angles

The angles 1 and 3 (180-α) and the angles 2 and 4 (α) are vertical angles. All pairs of vertical angles are equal.

 

The angles 1 and 2, the angles 2 and 3, the angles 3 and 4 and the angles 4 and 1 are supplementary angles. The sum of each pair of supplementary angles equals to 180 degrees (180-α+α=180°).

Consider the following example:

In the figure below are presented 4 lines that cross each other: AB, FI, AH and BG. The triangle ABC is an isosceles triangle, so that AS=BC. The angle ∠FEG is equal to 30° and the angle ∠FDH is equal to 120°.

What is the value of the angle BAC?

Angle relationships between intersecting lines

We are given that the angle ∠FEG is equal to 30°, therefore the angle ∠CED is also equal to 30° (the angles are vertical, therefore they are equal).

We are given that the angle ∠FDH is equal to 120°, therefore the angle ∠CDI is equal to 180°-120°=60° (the angles are supplementary, therefore their sum is 180°).

The sum of angles in a triangle is equal to 180°, therefore the sum of angles in the triangle CDE is 180° and the angle ∠DCE=180°-60°-30°=90°.

We are given that the triangle ABC is an isosceles triangle, therefore the angles ∠ABS and ∠BAC are equal.
The sum of angles in a triangle is equal to 180°, therefore the sum of angles in the triangle ABC is 180°.
We calculated that the angle ∠DCE is equal to 90° the angles ∠ABS and ∠BAC are equal.
From these 3 statements we can conclude that the angle ∠BAC=∠ABC=(180°-90°)/2=90°/2=45°.

Angle relationships between intersecting and parallel lines- alternate angles

Two parallel lines combined with two intersecting lines form 2 similar triangles and equal alternate angles.

Similar triangles have the same shape, but not the same size (they have the same angle measures, but not the same side lengths).
Note that the corresponding side lengths of similar triangles are related by a constant ratio, which is called k. See further details below.

Alternate angles are angles located in opposite positions relative to a transversal intersecting two parallel lines. Alternate angles are equal.

In the figure below two parallel lines AB and DE (marked in blue) combined with two intersecting lines AE and BD form 2 similar triangles ABC and CDE.

The angles x°, y° and z° are equal.

Angles between intersecting lines and parallel lines

The equal angles x°, y° and z°:

The equal angles z° are vertical angles that are formed by the lines AE and BD.

The equal angles x° are called alternate angles, they are formed by the parallel lines AB and DE and the line AE that crosses them.

The equal angles y° are called alternate angles, they are formed by the parallel lines AB and DE and the line BD that crosses them.

Since the angles x°, y° and z° are equal between the triangle ABC and CDE, the triangles are similar.  

Consider the following example:

In the figure below are presented 4 lines that cross each other: HI, DE, EF and BG. The lines HI and DE are parallel. The angle ∠FAH is equal to 40° and the angle ∠EDG is equal to α. In addition, AC=BC.

What is the value of α?

Angle relationships between intersecting lines and parallel lines question

We are given that the angle ∠EDG is equal to α.

Angles that lie on the same side of a straight line, are always supplementary and their sum is equal to 180°. Therefore, the angles ∠EDG and ∠CDE are supplementary and ∠EDG+ ∠CDE=180°.

The angle ∠CDE=180°-α.

Two parallel lines combined with two intersecting lines form 2 similar triangles. We are given that the lines HI and DE are parallel, therefore the triangles ABC and CDE are similar and the angles ∠CDE and ∠ABC are equal and their value is 180°-α (we found that ∠CDE=180°-α).

Note that instead of using similar triangles we can use the statement that two parallel lines combined with two intersecting lines form 2 equal alternate angles. We are given that the lines HI and DE are parallel, therefore the angles ∠CDE and ∠ABC are equal (these angles are alternate) and their value is 180°-α (we found that ∠CDE=180°-α).

We are given that the angle ∠FAH is equal to 40°, therefore the angle ∠BAC is also equal to 40° (the angles are vertical, therefore they are equal).

We are given that AC=BC, therefore the angle ∠BAC=∠ABC.

We found that the angle ∠BAC=∠ABC, the angle ∠BAC=40° and the angle ∠ABC=∠CDE=180°-α. Therefore, the angle ∠ABC=∠CDE=∠BAC=∠FAH=180°-α=40°

We can solve the equation 180°-α=40°, getting α=180°-40°=140°

Angle relationships between parallel lines inside a triangle- corresponding angles

A parallel line inside a triangle (the line is parallel to one of the sides of the triangle) forms 2 similar triangles (a small triangle and a large triangle) and equal corresponding angles.

Corresponding angles are angles located in the same position relative to parallel lines intersected by a transversal. Corresponding angles are equal.

In the figure below a parallel line EB inside a triangle ACD (EB is parallel to CD) forms 2 similar triangles (a small triangle ABE and a large triangle ACD).

 

Angle relationships between parallel lines inside a triangle- corresponding angles

The equal angles x° and y°:

The equal angles x° are called corresponding angles, they are formed by the parallel lines BE and CD and the line AD.

The equal angles y° are called corresponding angles, they are formed by the parallel lines BE and CD and the line AC.

Since the angles x°, y° and z° are equal between the triangle ABE and ACD, the triangles are similar.  

Consider the following example:

In the figure below are presented 4 lines that cross each other: AD,AC,BF and CE. The lines BF and CE are parallel. The angle ∠BAF is equal to 20° and the angle ADE is equal to α. In addition, AF=AB.

What is the value of α?

Angle relationships between parallel lines inside a triangle- corresponding angles

Angles that lie on the same side of a straight line, are always supplementary and their sum is equal to 180°. Therefore, the angles ADE and ADC are supplementary and ADE+ ADC=180°.

We are given that the angle ADE= α, therefore α+ADC=180° and the angle ADC=180°-α.

A parallel line inside a triangle forms 2 similar triangles. We are given that the lines BF and CE are parallel, therefore the triangles ABF and ACD are similar and the angles AFB and ADC are equal and their value is 180°-α (we found that ADC=180°-α).

Note that instead of using similar triangles we can use the statement that a parallel line inside a triangle forms 2 equal corresponding angles. We are given that the lines BF and CE are parallel, therefore the angles AFB and ADC are equal (these angles are corresponding) and their value is 180°-α (we found that ADC=180°-α).

We are given that AB=AF, therefore the angle ABF=AFB. We are also given that the angle BAF=20°.

The sum of angles in a triangle is equal to 180°, therefore the sum of angles in the triangle ABF is 180° and the angle ABF=AFB=(180°-20)/2=160°/2=80°.

We have an equation that we can solve: ABF=AFB=ADC=180°-α=80°.

180°-α=80°

α=180°-80°

α=100°

Note that the angle ACD=ABF=80°.

Calculating side lengths with triangles similarity

Similar triangles have the same shape, but not the same size (they have the same angle measures, but not the same side lengths).

The corresponding side lengths of similar triangles are related by a constant ratio, which is called k.

Note that:
Similar triangles have the same angle measures.
If 2 lines are parallel, the triangles that are formed by them are similar.

The figure below shows two similar triangles: ABC and DEF. The triangle ABC is similar to the triangle DEF.

Finding side lengths with triangles similarity

Similar triangles have the same angle measures, therefore the angle ∠A is equal to the angle ∠D; the angle ∠B is equal to the angle ∠E and the angle ∠C is equal to the angle ∠F.

The corresponding side lengths of similar triangles are related by a constant ratio, which is called k, therefore the following ratios between pairs of the sides exist:
DE=k*AB (marked in orange in the figure above)
DF=k*AC (marked in green in the figure above)
EF=k*BC (marked in black in the figure above)

We can also present all the ratios as being equal to k:

k =  DE   =   DF   =   EF
       ____     ____      ____
       AB        AC        BC

Note that we could write the ratio in an opposite way by switching between the numerator and the denominator (so that the numerator will include the sides of the small triangle and the denominator will include the sides of the big triangle). The parameter k will be smaller than 1 and equal to 1 divided by the k parameter above.
For example: if the ratio k between the sides of the big triangle divided by the sides of the small triangle is 2 (like 4 divided by 2), then the ratio k between the sided of the small triangle divided by the sides of the big triangle will be equal to 1/2 (like 2 divided by 4).

Consider the following example:

In the figure below is given that AB=5 centimeters; CD=27 centimeters; BE is parallel to CD; AC is 3 times bigger than AB.

What is the value of BE?

An example- Calculating side lengths with triangles similarity

We are given that AC is 3 times bigger than AB, therefore we can write an equation AC=AB*3.

We are given that AB=5 centimeters, therefore AC=AB*3=5*3=15. Meaning that AC/AB=3.

A parallel line inside a triangle forms 2 similar triangles. We are given that BE is parallel to CD, therefore the triangles ABE and ACD are similar.

The corresponding side lengths of similar triangles are related by a constant ratio, which is called k.

We found AC/AB=3, therefore k=3.

K is also equal to BE/CD, therefore we can write an equation CD/BE=3.

We are given that CD=27 centimeters, therefore we can write an equation 27/BE=3.

27/BE=3

BE=27/3

BE=9 centimeters

Calculating side lengths with triangles similarity when one side is common for both triangles

In these questions one side is common for both triangles, therefore we need the length of two sides instead of three sides to calculate the length of the fourth side using the similar triangles ratio.  

Consider the following example:

The figure below presents two similar triangles, ABC and BDC.

The length of AB is 20 centimeters and the length of BC is 15 centimeters.

What is the length of DC?

Example -calculating side lengths with triangles similarity when one side is common for both triangles

In this example the side BC is common for both the triangles. Therefore we can you it twice in the similar triangles ratio.

The similar triangles ratio is

k   =   AB     =    BC
          ____        ____
           BC           DC

k   =   20   =  15
         ____     ____
          15        DC

4  =  15
__      __
3      DC

4DC=45

DC=45/4=11 1/4

The length of the side DC is 11 1/4 centimeters.

Note that we could write the ratio in an opposite way by switching between the numerator and the denominator and get the same answer:

k    =   BC     =    DC
           ____         ____
           AB            BC

K   =   15   =    DC
         _____      _____
           20          15

 3   =   DC
___      ___
 4         15

4DC=45

DC=45/4=11 1/4

The ratio k in this case will be 15/20=3/4, which is 1 divided by the previous ratio k=4/3.

You just finished studying congruence and similarity topic, the third topic of additional topics in math!

Continue studying the next additional topic in math- right triangle trigonometry and word problems.

Right triangle trigonometry and word problems

Right triangle trigonometry and word problems on the SAT test

SAT Subscore: Additional topics in math

Studying right triangle trigonometry and word problems

On the SAT test right triangle trigonometry and word problems topic is the fourth topic of additional topics in math that include 7 advanced topics (see the full topics list on the top menu). It is recommended to start learning additional topics in math with its first topic called complex numbers.

Right triangle trigonometry and word problems topic is divided into sections from easy to difficult (the list of the sections appears on the left menu). Each section includes detailed explanations of the required material with examples followed by a variety of self-practice questions with solutions.

Finish studying heart of algebra subscore topics before you study this topic or any other additional topic in math. (Heart of algebra subscore includes basic algebra topics which knowledge is required for understanding additional topics in math). 

Right triangle trigonometry and word problems- summary

Right triangle is a triangle with a right angle (equal to 90°). The side opposite the right angle (the longest side of the right triangle) is called a hypotenuse.

Right triangle trigonometry and right triangle word problems require calculating side lengths and angle measures in right triangles.

Pythagorean theorem:

Pythagorean theorem states that the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides: a2 + b2 = c2

Pythagorean triples are combinations of side lengths a, b and c that satisfy the Pythagorean theorem. If you remember the triples values, you know the size of the third side without the need to calculate it. The most common Pythagorean triples are: 3, 4 and 5;    5, 12 and 13;    7, 24 and 25.  

Trigonometric ratios (sine, cosine and tangent):

Trigonometric ratios (functions) represent connections between angle degrees and side lengths in a right triangle:

  • The sine of an angle (sin) in a right triangle is defined as the ratio of the length of the side that is opposite to the angle, to the length of the hypotenuse. 
  • The cosine of an angle (cos) in a right triangle is defined as the ratio of the length of the side that is adjacent to the angle, to the length of the hypotenuse. 
  • The tangent of an angle (tan) in a right triangle is defined as the ratio of the length of the side that is opposite to the angle, side that is adjacent to the angle. 

Complementary angles are two angles with the sum of 90 degrees. Sine of an angle (α) in a right triangle is equal to cosine of its complementary angle (90-α). 

Similar triangles have the same angle measures and their corresponding side lengths are related by a constant ratio therefore they also have similar sine, cosine and tangent.

Special right triangles:

Special right triangles are right triangles whose sides are in a particular ratio. 

  • In a 30°, 60°, 90° right triangle the side opposite the 30° angle is half the length of the hypotenuse and the side opposite to 60° angle is equal to the length of a side opposite to 30° angle multiplied by √3.
    In 30°, 60°, 90° triangle the sides are x, x√3 and 2x.
  • In a 45°, 45°, 90° right triangle the sides opposite the 45° angles are equal and the hypotenuse is equal to the side opposite to 45° angle multiplied by √2.
    In 45°, 45°, 90° triangle the sides are s, s and s√2.

 

Continue reading this page for detailed explanations and examples.

Pythagorean theorem in right triangles

Pythagorean theorem states that the square of the length of the hypotenuse of a right triangle equals the sum of the squares of the lengths of the other two sides. (The hypotenuse is the side opposite the right angle).

The formula of the Pythagorean theorem is a2 + b2 = c2. This formula is provided at the beginning of each math section followed by a diagram.  

The following figure presents a right triangle and the Pythagorean theorem a2 + b2 = c2.

Pythagorean theorem in a triangle

Calculating side lengths with Pythagorean theorem

If we are given the values of two sides of a right triangle, we can calculate the value of the hypotenuse with the Pythagorean theorem.

If we are given the values of one side and the hypotenuse of a right triangle, we can calculate the value of the second side with the Pythagorean theorem.

Consider the following example:

The length of the sides of a right triangle are 3 centimeters and 4 centimeters.

What is the length of the hypotenuse?

The formula of the Pythagorean theorem is a2+b2=c2.
a=3, b=4, c=?
32 + 42 = c2
c2=9+16
c2=25
c=5
The length of the hypotenuse is 5 centimeters.

Checking the answer:
32+42=52
9+16=25
25=25

Consider the following example:

The length of a side of a right triangle is 4 centimeters and the length of the hypotenuse is 8 centimeters.

What is the length of the other side?

The formula of the Pythagorean theorem is a2+b2=c2.
a=4, b=?, c=8
42+b2=82
16+b2=64
b2=64-16
b2=48
b=√48
b=√(4*12)
b=√4*√12
b=2√12=6.93

Checking the answer:
a2+b2=c2
42+(2√12)2=82
16+4*12=64
16+48=64
64=64

Pythagorean triples in a right triangle

Pythagorean triples are combinations of side lengths a, b and c that satisfy the Pythagorean theorem. If you remember the triples values, you know the size of the third side without the need to calculate it.

The most common Pythagorean triple is 3, 4 and 5.
Additional Pythagorean triples are:
5, 12 and 13
7, 24 and 25  

Note that each multiplication of the triple also satisfies the Pythagorean theorem.
For example: The triple 9, 12 and 15 is the triple 3, 4 and 5 multiplies by 3. Therefore, the triple 9, 12 and 15 is a Pythagorean triple.
We can check this statement by plugging into the formula of the Pythagorean theorem:
a2+b2=c2
92+122=152
81+144=225
225=225

Consider the following example:

The length of a side of a right triangle is 7 centimeters and the length of the hypotenuse is 25 centimeters.

What is the length of the other side?

The triple 7, 24 and 25 is a Pythagorean triple, therefore the length of the second side is 24 centimeters.

Checking the answer:
a2+b2=c2
72+b2=252
b2=252-72
b2=625-49
b2=576
b=24
The length of the other side is 24 centimeters.

Consider the following example:

The length of the sides of a right triangle are 10 centimeters and 24 centimeters.

What is the length of the hypotenuse?

The triple 5, 12 and 13 is a Pythagorean triple, the given sides 10 and 24 are the triple sides 5 and 12 multiplied by 2. Therefore, the length of the hypotenuse is 13 multiplied by 2=26 centimeters.

Checking the answer:
a2+b2=c2
102+242=262
100+576=676
676=676 

Calculating triangles side lengths with Pythagorean theorem and triangles similarity

Similar triangles have the same shape, but not the same size (they have the same angle measures, but not the same side lengths).

The corresponding side lengths of similar triangles are related by a constant ratio, which is called k.

 Note that:
Similar triangles have the same angle measures.
If 2 lines are parallel, the triangles that are formed by them are similar.

Given parallel lines and a right triangle, we can calculate side lengths using triangles similarity combined with the Pythagorean theorem.

Calculating side ratio with triangles similarity

Consider the following example:

In the figure below the lines BE and CD are parallel. The angle measures and the side lengths (in centimeters) are given in the figure.

What are the lengths of the sides of triangle ACD?

Calculating right triangles side lengths

Calculating the length of AB:

Since BE is parallel to CD, the triangles ABE and ACD are similar. Therefore, the angle ACD is equal to the angle ABE and equal to 90 degrees.

Since the angle ABE is equal to 90 degrees, the triangle ABE is a right triangle. Therefore, we can apply the Pythagorean theorem in the triangle ABE:

The sides 3, 4 and 5 are Pythagorean triple, or we can solve the Pythagorean theorem:
a2+b2=c2
32+42=c2
c2=9+16
c2=25
c=5
AB=5 centimeters

Calculating the length of CD and DE:
AC=BC+AB
AC=5+5=10 centimeters

Since the lines BE and CD are parallel, the triangles ACD and ABE are similar. Therefore, the ratio of the related sides is k=AC/AB=10/5=2 (the side of the big triangle divided by the side of the small triangle is equal to 2).

Since k=2, AD/AE=2 and CD/BE=2.

Since AE=4 and k=2, AD/4=2, AD=8 centimeters and DE=AD-AE=8-4=4 centimeters.
Since BE=3 and k=2, CD/3=2, CD=6 centimeters.

Checking by calculating Pythagorean theorem in the big triangle ACD:
a2+b2=c2
62+82=102
This is a Pythagorean triple 3, 4 and 5 multiplied by 2.

Trigonometric ratios (sine, cosine and tangent)

Trigonometric ratios (functions) represent connections between angle degrees and side lengths in a right triangle.

The sine of an angle (sin) in a right triangle is defined as the ratio of the length of the side that is opposite to the angle, to the length of the hypotenuse. In the figure below in a right triangle ABC, sine (A)= BC/AC.

The cosine of an angle (cos) in a right triangle is defined as the ratio of the length of the side that is adjacent to the angle, to the length of the hypotenuse. In the figure below in a right triangle ABC, cosine (A)= AB/AC.

The tangent of an angle (tan) in a right triangle is defined as the ratio of the length of the side that is opposite to the angle, side that is adjacent to the angle. In the figure below in a right triangle ABC, tangent (A)= BC/AB.

In the figure below we see a right triangle ABC.
sin (A)=opposite/hypotenuse=BC/AC
cos (A)=adjacent/hypotenuse=AB/AC
tan (A)=opposite/adjacent=BC/AB

Trigonometric functions: sine, cosine, tangent

Sine and cosine of complementary angles

Complementary angles are two angles with the sum of 90 degrees.

Complementary angles in a right triangle: Since one angle in a right triangle is equal to 90 degrees and the sum of angles of a triangle is 180 degrees, the two other acute angles are complementary.

Sine of an angle in a right triangle is equal to cosine of its complementary angle. Meaning that given two complementary angles α and 90-α in a right triangle, sinα=cos(90-α).

Showing the connection between sine and cosine of complementary angles:
If we represent by α the measure of one angle in a right triangle, then the other angle measure is 90-α (complementary angles).

The figure below presents a right triangle ABC with complementary angles α and 90-α.

sinα=BC/AC

cos(90-α)=BC/AC

We see that sinα=cos(90-α)

Sine and cosine equality of complementary angles

Consider the following example:

What is sin (60) is equal to:
A. cos(30)
B. tan(60)
C. sin(30)
D. cos(60)

We know that sinα=cos(90-α), therefore, sin60=cos(90-60)=cos(30).
The answer A is correct.

Calculation of side lengths with sine, cosine and tangent

To calculate a side length, we need to know the value of the trigonometric function (sine, cosine or tangent) and the value of the other side.

Consider the following example:

The figure above represents a right triangle ABC.

If sin(a)= 0.5 and AC=5 centimeters, what is the value of AC?

sin(A)=BC/AC

0.5=BC/5

BC=0.5*5=2.5 centimeters

Calculation of sine, cosine and tangent in similar triangles

Similar triangles have the same angle measures and their corresponding side lengths are related by a constant ratio therefore they also have similar sine, cosine and tangent.

We can calculate sine, cosine or tangent in one triangle and conclude that they are identical in all similar triangles.

Related corresponding side lengths lead to similar sine, cosine and tangent:
If the sides of triangle 1 are x, y and z and the sides of a similar triangle 2 are kx,ky and kz (multiplied by a factor of k) then:
The sine of an angle A in triangle 1 is sin A1=x/z
The sine of an angle A in triangle 2 is sin A2=kx/kx=x/z
The sine of triangle 1 is equal to the sine of triangle 2.

Consider the following example:

In the figure below the lines BE and CD are parallel. The angle measures and the side lengths (in centimeters) are given in the figure.

What is the cosine of angle E?

Sine, cosine and tangent in similar triangles

Special right triangles

Special right triangles are right triangles whose sides are in a particular ratio. Two special right triangles are 30°, 60°, 90° triangle and 45°, 45°, 90° triangle.

The special right triangles with the side sizes length are given at the beginning of each SAT section.

30°, 60°, 90° triangle:
In a 30°, 60°, 90° right triangle the side opposite the 30° angle is half the length of the hypotenuse and the side opposite to 60° angle is equal to the length of a side opposite to 30° angle multiplied by √3.
In 30°, 60°, 90° triangle the sides are x, x√3 and 2x.

 45°, 45°, 90° triangle:
In a 45°, 45°, 90° right triangle the sides opposite the 45° angles are equal and the hypotenuse is equal to the side opposite to 45° angle multiplied by √2.
In 45°, 45°, 90° triangle the sides are s, s and s√2.

The following graphs present the special right triangles with the side sizes length.

Two special right triangles are 30°, 60°, 90° triangle and 45°, 45°, 90° triangle.

Calculation of side lengths given angle measures in special triangles

Given the length of any side in a special right triangle, we can calculate the length of the two other sides.

If you identify that the angles of a right triangle have the measures of 30,60,90 or 45,45,90 you know the ratios between their side length according to the graph presented above:

In 30°, 60°, 90° triangle the sides are x, x√3 and 2x.

In 45°, 45°, 90° triangle the sides are s, s and s√2.

 

Consider the following example:

In a right triangle the measures of the angles are 30°, 60°, 90° and the hypotenuse is equal to 8 centimeters.

What are the lengths of the other sides?

In 30°, 60°, 90° triangle the sides are x, x√3 and 2x.

Given the hypotenuse is equal to 8 centimeters, we know that the side opposite to 30° angle is equal to 8/2=4 centimeters. The side opposite to 60° angle is equal to 4*√3=√16*√3=√48.

Checking with Pythagorean theorem:
a2+b2=c2
42+√482=82
16+48=64
64=64

Consider the following example:

In a right triangle the measures of the angles are 45°, 45°, 90° and the hypotenuse is equal to √40 centimeters.

What are the lengths of the sides?

In 45°, 45°, 90° triangle the sides are s, s and s√2.

We are given that the hypotenuse is equal to √40 centimeters, therefore s√2=√40. Continue solving:
s√2=√40
s=√40/√2
s=√20

Checking with Pythagorean theorem:
a2+b2=c2
√202+√202=√402
20+20=40
40=40

Calculation of angle measures given side lengths in special triangles

Given the side lengths of a special right triangle, we can calculate the angle measures.

 If the ratio between a side and a hypotenuse in a right triangle is 1/2 we know that the triangle is a special right triangle with 30°, 60°, 90° angles (the side used in the ratio is located opposite to the 30° angle).

If the ratio between a side and a hypotenuse in a right triangle is √3/2 we know that the triangle is a special right triangle with 30°, 60°, 90° angles (the side used in the ratio is located opposite to the 60° angle).

If the ratio between two sides in a right triangle is √3 or 1/√3 we know that the triangle is a special right triangle with 30°, 60°, 90° angles (the bigger side which is multiplied by √3 is located opposite to the 60° angle).

If a right triangle is an isosceles triangle, it is a special right triangle with angles 45°, 45° and 90°.

If the ratio between a side and a hypotenuse in a right triangle is 1/√2 we know that the triangle is a special right triangle with 45°, 45°, 90° angles.

Consider the following example:

The ratio between two sides in a right triangle is √27/9, what are the measures of the angles of the triangle?

Simplifying the ratio √27/9 gives us:
√27/9=√27/√81
√27/√81=√27/(√27*√3)
√27/(√27*√3)=1/√3

If the ratio between two sides in a right triangle is 1/√3 we know that the triangle is a special right triangle with 30°, 60°, 90° angles (the bigger side which is multiplied by √3 is located opposite to the 60° angle).

You just finished studying right triangle trigonometry and word problems topic, the fourth topic of additional topics in math!

Continue studying the next additional topic in math- circle theorems.

Volume word problems

Volume word problems on the SAT test

SAT Subscore: Additional topics in math

Studying volume word problems

On the SAT test volume word problems topic is part of additional topics in math that include 7 advanced topics (see the full topics list on the top menu). 

Volume word problems topic is the second topic of additional topics in math. It is recommended to start learning additional topics in math with its first topic called complex numbers.

Volume word problems topic is divided into sections from easy to difficult (the list of the sections appears on the left menu). Each section includes detailed explanations of the required material with examples followed by a variety of self-practice questions with solutions.

Finish studying heart of algebra subscore topics before you study this topic or any other additional topics in math. (Heart of algebra subscore includes basic algebra topics which knowledge is required for understanding additional topics in math). 

Volume word problems- summary

Volume word problems require making calculations of volumes of three-dimensional shapes using volume formulas. To calculate the volume, plug the given dimension into the relevant volume formula.

Volume formulas of five basic shapes:
The volume formulas of five basic shapes are given at the beginning of the math sections of the SAT exam: right rectangular prism, right circular cylinder, sphere, right circular cone and rectangular pyramid.
Right rectangular prism volume formula is V= lwh.
Right circular cylinder volume formula is V=πr2h.
A sphere volume formula is V= 4/3 πr3.
A right circular cone volume formula is V= 1/3 πr2h.
A rectangular pyramid volume formula is V=1/3 lwh.

Calculating the effect of changes in dimensions on volume: The power of the dimension determines the size of the change in the volume value.

  • If a dimension in the volume formula is raised to a first power, the volume changes by the same factor as the shape.
  • If a dimension in the volume formula is raised to a second power, when the shape changes by a factor the volume changes by a square of the factor.
  • If a dimension in the volume formula is raised to a third power, when the shape changes by a factor the volume changes by a third degree of the factor.

Comparing volumes of two shapes: In these questions we are given ratios between the dimensions of two shapes and we are required to compare their volumes. We need to calculate the total effect on the volume of all the ratios between the shapes. 

Continue reading this page for detailed explanations and examples.

Volume formulas worksheet

Volume formulas worksheet is given below, it includes 5 formulas:

Right rectangular prism volume formula is V= lwh.

Right circular cylinder volume formula is V=πr2h.

A sphere volume formula is V= 4/3 πr3.

A right circular cone volume formula is V= 1/3 πr2h.

A rectangular pyramid volume formula is V=1/3 lwh.

In the formulas below:
l=length, w=width, h=height, V=volume, A=area
Circle measures: π=3.14159, diameter=2*radius

Right rectangular prism volume formula, Right circular cylinder volume formula, A sphere volume formula, A right circular cone volume formula, A rectangular pyramid volume formula

The volume of a right rectangular prism

A right rectangular prism is a three-dimensional object with 6 faces, where all the 6 faces are rectangles. In a right rectangular prism, the angles between the base and sides are right angles.

The volume of a right rectangular prism is defined as the product of the area of one face (length*width) multiplied by its height.

Right rectangular prism volume formula is V= lwh.

Consider the following example:

Calculate the volume of a rectangular prism that has a height of 6 centimeters, a length of 3 centimeters and a width of 4 centimeters.

Right rectangular prism volume formula is V= lwh.

V=lwh=3*4*6=72 cubic centimeters.

The volume of a right circular cylinder

A right circular cylinder has 2 identical and parallel circular bases at the ends. The elements are perpendicular to the bases, therefore the cylinder is called right.

The volume of a right circular cylinder is defined as the product of the area of the circular base multiplied by the height of the cylinder.

Circular base area formula is A=πr2.

Right circular cylinder volume formula is V=πr2h.

Consider the following example:

Calculate the volume of a circular cylinder that has a radius of 5 centimeters and a height of 10 centimeters.

Right circular cylinder volume formula is V=πr2h.

V=πr2h=π*52*10=250π=785 cubic centimeters.

The volume of a sphere

A sphere is a three-dimensional object that has a surface of a ball, all the point on the surface of the sphere are lying at the same distance (the radius) from the center.

A sphere volume formula is V= 4/3 πr3.

Consider the following example:

Calculate the volume of a sphere that has a radius of 5 centimeters.

 A sphere volume formula is 4/3 πr3.

V=4/3 πr3=4/3*π*53= 4/3*53*π=167π =524 cubic centimeters.

The volume of a right circular cone

A right circular cone is a cone in which the center point of the circular base is joined with the vertex of the cone and forms a right angle (the height is perpendicular to the radius or the circle).

A right circular cone volume formula is V= 1/3 πr2h.

Consider the following example:

A cone has a height of 10 centimeters and a circular base with a radius of 3 centimeters. What is the volume of the cone?

A right circular cone volume formula is V= 1/3 πr2h.

V= 1/3 πr2h=1/3*π*32*10=30π=30*3.14159=94 cubic centimeters.

The volume of a rectangular pyramid

A rectangular pyramid is a pyramid that has four-sided base and a vertex.

A rectangular pyramid volume is defined as the product of the area of the base multiplied by the height of the pyramid (the height is the distance from the center point of the base to the vertex) divided by 3.

A rectangular pyramid volume formula is V=1/3 lwh.

Consider the following example:

A pyramid has a height of 10 centimeters a length of 3 centimeters and a width of 5 centimeters. What is the volume of the pyramid?

A rectangular pyramid volume formula is V=1/3 lwh.

V=1/3 lwh= 1/3*3*5*10=50 cubic centimeters.

Calculating dimensions given the volumes values of the shapes

To calculate an unknown dimension, plug the given dimensions and the volume values into the volume formula and solve.

Consider the following example:

A sphere has a volume of 33.5 cubic centimeters.

What is the radius of the sphere?

A sphere volume formula is V= 4/3 πr3.
4/3 πr3=33.5
πr3=33.5*3/4
πr3=25.125
3.14159r3=25.125
r3=25.125/3.14159
r3=8
r=3√8
r=2
The radius of the sphere is 2 centimeters.

Consider the following example:

A pyramid has a square base with a height that is 3 timed bigger than the length of the base.

If the volume of the pyramid is 27 cubic centimeters, what is its height?

x= the length and the width of the square pyramid
3x= the height of the pyramid

A rectangular pyramid volume formula is V=1/3 lwh.
V=1/3 lwh= 1/3 x*x*3x
1/3 x*x*3x=27
x3=27
x=3√27
x=3

Calculating the effect of changes in dimensions on volume

The power of the dimension determines the size of the change in the volume value:

If a dimension in the volume formula is raised to a first power, the volume changes by the same factor as the shape.
For example: A rectangular pyramid volume formula is V=1/3 lwh. The length, width and height in the formula are raised to a first power. If we double the length or the width or the height (one of them), then the volume will also be doubled.

If a dimension in the volume formula is raised to a second power, when the shape changes by a factor the volume changes by a square of the factor.
For example: A right circular cone volume formula is V= 1/3 πr2h. The radius of the circle is raised to a second power. If we double the radius, the volume will change by 22=4. The height in the formula is raised to a first power. If we double the height, then the volume will also be doubled.

If a dimension in the volume formula is raised to a third power, when the shape changes by a factor the volume changes by a third degree of the factor.
For example: A sphere volume formula is 4/3 πr3. The radius of the sphere is raised to a third power. If we double the radius, the volume will change by 23=8.

Consider the following example:

A circular cylinder has a radius of 5 centimeters and a height of 10 centimeters. What is the change in the volume if we double the radius and halve the height?

Right circular cylinder volume formula is V=πr2h.

We see in the formula that the radius is raised to a second power and the height is raised to a first power.
If we doble the radius, the volume will be multiplied by 22=4.
If we halve the height, the volume will be multiplied by 0.51.
The total change in the volume is 4*0.5=2.

We can check the answer by calculating the volumes in the two scenarios:
V=πr2h=π*52*10= 250π cubic centimeters.
V=πr2h=π*102*5= 500π cubic centimeters.
The volume value is doubled.

Comparing volumes of two shapes

In these questions we are given ratios between the dimensions of two shapes and we are required to compare their volumes.

We need to calculate the total effect on the volume of all the ratios between the shapes.

Consider the following example:  

A right circular cone has a volume of 200 cubic centimeters.
A right circular cone volume formula is V= 1/3 πr2h.

Which of the following cones has a higher volume than the given cone?
A. A cone with 3 times the radius and 1/10 times the height.
B. A cone with 2 times the radius and 1/4 times the height.
C. A cone with 1/2 times the radius and 3 times the height.
D. A cone with 1/3 times the radius and 10 times the height.

A right circular cone volume formula is V= 1/3 πr2h.

We see in the formula that the radius is raised to a second power and the height is raised to a first power.

Answer A: A cone with 3 times the radius and 1/10 times the height:
The total change is (32)*1/10=9/10 <1.
This cone has a smaller volume than the given cone.

Answer B: A cone with 2 times the radius and 1/4 times the height.
The total change is (22)*1/4=4/4=1
This cone has the same volume as the given cone.

Answer C: A cone with 1/2 times the radius and 3 times the height.
The total change is (1/2)2*3=(1/4)*3=2/4 <1.
This cone has a smaller volume than the given cone.

Answer D: A cone with 1/3 times the radius and 10 times the height.
The total change is (1/3)2*10=(1/9)*10=10/9 >1
This cone has a bigger volume than the given cone.

The correct answer is D– A cone with 1/3 times the radius and 10 times the height has a bigger volume than the given cone.

Volume formulas of five basic shapes

Right rectangular prism volume formula, Right circular cylinder volume formula, A sphere volume formula, A right circular cone volume formula, A rectangular pyramid volume formula

You just finished studying volume word problems topic, the second topic of additional topics in math!

Continue studying the next additional topic in math- congruence and similarity.

Center, spread and shape of distributions

Center, spread and shape of distributions on the SAT test

SAT Subscore: Problem solving and data analysis

Studying center, spread and shape of distributions

On the SAT test center, spread and shape of distributions topic is part of problem solving and data analysis subscore that includes 9 advanced topics (see the full topics list on the top menu). 

Center, spread and shape of distributions topic is the seventh topic of problem solving and data analysis subscore. It is recommended to start learning problem solving and data analysis subscore with its first topic called ratios, rates and proportions.

Center, spread and shape of distributions topic is divided into sections from easy to difficult (the list of the sections appears on the left menu). Each section includes detailed explanations of the required material with examples followed by a variety of self-practice questions with solutions.

Finish studying heart of algebra subscore topics before you study this topic or any other problem solving and data analysis subscore topic. (Heart of algebra subscore includes basic algebra topics which knowledge is required for understanding problem solving and data analysis subscore topics). 

Center, spread and shape of distributions- summary

Center, spread and shape of distributions are statistical measures that describe data sets, they are called summary statistics.

A center of a data set is a way of describing a location. We can measure a center of a data in 3 different ways: the mean (average), the median and the mode.

A spread of a data set describes how similar or varied the set of the observed values. We can measure a center of a data in 2 different ways: a range and a standard deviation.

Center measures

The mean is the average value of a given data set.
Mean = average = sum of the values / number of the values

The median is the middle number in a sorted in ascending order data set (the median is the value that splits the data set into two halves). To calculate the median,  arrange the values in an ascending order, count them and calculate the median. If the number of values is odd, the median is the middle value. If the number of values is odd, the median is the average of the two middle values.

The mode of a data set is the number that occurs most frequently in the set. To determine the mode, order the numbers from least to greatest, count how many times each number occurs and determine the mode. If no value appears more than once in the data set, the data set has no mode. If a there are two values that appear in the data set an equal number of times, they both will be modes etc.

Spread measures

The range measures the spread of a data inside the limits of a data set, it is calculated as a difference between the highest and lowest values in the data set. The larger the range, the greater the spread of the data.
range= the highest value – the lowest value.

The standard deviation is the measure of the overall spread (variability) of a data set values from the mean. The spread is measured as the distances (absolute values) from the mean of each value of the data sat. The more spread out a data set is, the greater are the distances from the mean and the standard deviation.

Outliers

An outlier is a value that is very different from the other values, so that it lies an abnormal distance from other values and is far from the middle of the data set.

  • Mean: Removing a big outlier will reduce the mean value and removing a small outlier will enlarge the mean value. 
  • Median: Removing a small outlier will enlarge the median; removing a large outlier will reduce the median (the median will be the same if the values in the positions after the removal are equal to the values in the positions before the removal).
  • Range: The value that will replace the outlier will be less distant therefore the range after removing the outlier will be smaller.
  • Standard deviation: Since the outlier is a value that is far from other values and the mean, its removal will reduce the spread of the data and the standard deviation.  

 

Continue reading this page for detailed explanations and examples.

Measuring a center of a data set

A center of a data set is a way of describing a location. We can measure a center of a data in 3 different ways: the mean (average), the median and the mode.

We can measure a center of a data in 3 different ways: the mean (average), the median and the mode.

Mean calculation

Mean is an average value of a given data set. To calculate the mean, we need to add the total values given in a data set and then divide the sum by the total number of the values.

The mean formula is:
Mean = average = sum of the values / number of the values

Note that if a single value appears in a data set number of times, we need to include it number of times when calculating the sum of the values.

Consider the following example:

What is the mean (average) of the following numbers 10, 12, 16, 5 and 2?

mean = sum of the values / number of the values

mean=(10+12+16+5+2)/5=45/5=9.

Consider the following example:

There are 3 children in 3 families and 2 children in 2 families.

What is the mean (average) of the children in a family?

We need to translate the word problem into numerical values. Since the values 2 and 3 appears in a data set number of times, we need to include them number of times when calculating the sum of the values.

The values of the number of children in the families are 3, 3, 3, 2 and 2.

mean = sum of the values / number of the values

mean=(3+3+3+2+2)/5=13/5=2.6 children in a family.

Finding missing values given the mean

The mean formula is:
Mean = average = sum of the values / number of the values

If we are given the mean value, we can solve the equation of the mean formula for 1 value that is missing. This value can be one of the numbers in the data set.

For example:
In a data set {-21, 5, x, 10} the mean is 0.5.  The value of x is:
Mean = sum of the values / number of the values
(-21+5+x+10)/4=0.5
-21+5+x+10=4*0.5
-6+x=2
x=8

Consider the following example:

The average grade of 4 students is 75.

The teacher added another grade so that the average became 5 points higher.

What was the grade that the teacher added?

average = sum of the values / number of the values

(4*75+x)/5=75+5
300+x=80*5
300+x=400
x=400-300
x=85

Checking the answer: (4*75+100)/5=(300+100)/5=400/5=80.

Median calculation

The median is the middle number in a sorted in ascending or descending order data set. In other words, the median is the value that splits the data set into two halves.

Medial calculation steps:

Step 1: Arrange the values in an ascending (or a descending) order.

Step 2: Count how many values are in the data set.

Step 3: Calculate the median: If the number of values is odd, the median is the middle value. If the number of values is odd, the median is the average of the two middle values.

Note that if a single value appears in a data set number of times, we need to include it number of times when arranging the values in an ascending (or a descending) order.

Consider the following example:

What is the median of the following numbers 10, 12, 16, 5 and 2?

Step 1: Arranging the values in an ascending order: 2, 5, 10, 12 and 16.

Step 2: Counting how many values are in the data set: there are 5 values in the data set.

Step 3: Calculating the median: the number of values is odd, therefore the median is the middle value which is 10.

2, 5, 10, 12, 16

Consider the following example:

There are 3 children in 3 families, 1 child in 3 families and 2 children in 2 families.

What is the median of the children in a family?

We need to translate the word problem into numerical values. Since the values 1, 2 and 3 appears in a data set number of times, we need to include them number of times when arranging the values in an ascending order.

The values of the number of children in the families are 3, 3, 3, 1, 1, 1, 2 and 2.

Step 1: Arranging the values in an ascending order: 1, 1, 1, 2, 2, 3, 3 and 3.

Step 2: Counting how many values are in the data set: there are 8 values in the data set.

Step 3: Calculating the median: the number of values is even, therefore the median is the average of the two middle values in the locations 4 and 5 which is (2+2)/2=2.

1, 1, 1, 2, 2, 3, 3, 3

Calculating a median of a frequency graph

Arrange the data given in the graph in a table and follow the previous steps.

Consider the following example:

The table below shows exam grades of a group of students.

What is the median grade?

a median value of a frequency graph

Arranging the data from the graph in a table:
65-70      3
70-75      1
75-80      2
80-85      0
85-90      3
90-95      1

Arranging the values in an ascending order: 65-70, 65-70, 65-70, 70-75, 75-80, 75-80, 85-90, 85-90, 85-90, 90-95.

The number of the values is 3+1+2+3+1=10. The number of the values is even, therefore the median is the average of the groups in the positions 5 and 6. This is the group 75-80.

65-70, 65-70, 65-70, 70-75, 75-80, 75-80, 85-90, 85-90, 85-90, 90-95

Mode calculation

The mode of a data set is the number that occurs most frequently in the set.

Calculating the mode steps:

Step 1: Order the numbers from least to greatest.

Step 2: Count how many times each number occurs.

Step 3: Determine the mode- a data set can have more than one mode or no mode:

If no value appears more than once in the data set, the data set has no mode.

If a there are two values that appear in the data set an equal number of times, they both will be modes etc.

Consider the following example:

What is the mode of the following numbers 10, 12, 16, 5 and 2?

Step 1: Ordering the numbers from least to greatest- 2, 5, 10, 12 and 16.

Step 2: Counting how many times each number occurs- each number occurs one time.

Step 3: Determining the mode- there is no mode in the data set.

Consider the following example:

There are 3 children in 3 families, 1 child in 3 families and 2 children in 2 families. What is the mode of the children in a family?

This word problem already describes the values in groups, therefore we don’t need to arrange and count the values. Since 3 is the largest number of families, there are two modes which are 1 and 3 children.

1, 1, 1, 2, 2, 3, 3, 3  

Measuring a spread of a data set

A spread of a data set describes how similar or varied the set of the observed values.

We can measure a center of a data in 2 different ways: a range and a standard deviation.

Range calculation

A range measures the spread of a data inside the limits of a data set, it is calculated as a difference between the highest and lowest values in the data set. The larger the range, the greater the spread of the data.

A range formula is: range= the highest value – the lowest value.

Consider the following example:

What is the range of the following numbers 10, 12, 16, 5 and 2?

range= the highest value – the lowest value

range=16-2=14.

Consider the following example:

There are 3 children in 3 families, 1 child in 3 families and 2 children in 2 families. What is the range of the number of children in a family?

range= the highest value – the lowest value

range=3-1=2.

Standard deviation calculation

Standard deviation is the measure of the overall spread (variability) of a data set values from the mean. The spread is measured as the distances (absolute values) from the mean of each value of the data sat. The more spread out a data set is, the greater are the distances from the mean and the standard deviation.

Standard deviation calculation is not covered in the SAT, but you need to know to determine which data group has a greatest standard deviation.

Comparing standard deviations of number of data sets steps:

Step 1: Calculate the mean of each data set.

Step 2: Calculate the distance of each value from the mean. Note that the distance should be calculated as an absolute value.

Step 3: Summarize the distances (absolute values from step 2) of each dataset. The data set that has the smaller sum has the smallest standard deviation.

Consider the following example:

Which of the two following groups of numbers has a higher standard deviation?

Group A: 12, 16 and 20.

Group A: 7, 10 and 16.

Step 1: Calculating the mean of each group:
Group A: mean = average = sum of the values / number of the values=(12+16+20)/3=48/3=16.
Group B: mean = average = sum of the values / number of the values =(7+10+16)/3=33/3=11.

Step 2: Calculating the distance of each value from the mean:
Group A: 12-16=-4, 16-16=0, 20-16=4.
Group B: 7-11=-4, 10-11=-1, 16-11=5.

Step 3: Summarizing the distances (absolute values from step 2) of each group:
Group A: 4+0+4=8.
Group B: 4+1+5=10.

Group B has a bigger standard deviation than group A.

Outliers and their effect on summary statistics measures

An outlier is a value that is very different from the other values, so that it lies an abnormal distance from other values and is far from the middle of the data set.

For example:
In a data set of 5, 8, 10 and 30 the outlier is 30.
In a data set -21, 5, 8, 10 the outlier is -21.

The effect of an outlier on a mean

The mean formula is: mean = average = sum of the values / number of the values

Removing a value from a data set reduces the number of values (denominator) by 1 and changes the sum of the values (numerator). If we remove a value that is smaller than the mean, the new mean will be bigger; If we remove a value that is bigger than the mean, the new mean will be smaller; if we remove a value that is equal to the mean, the new mean will be without change. We know that the outlier is significantly smaller or bigger than the mean, therefore removing a big outlier will reduce the mean value and removing a small outlier will enlarge the mean value.

Consider the following example:

Calculate the mean of the following data sets with and without the outlier:

{5, 7, 8, 40}

{-21, 5, 8, 10}

In a data set of {5, 7, 8, 40} the mean is (5+7+8+40)/4=60/4=15.
The data set without the outlier of 40 is {5, 7, 8}, its mean is (5+7+8)/3=20/3=6.333.
The mean reduced from 15 to 6.33, since the outlier 40 is larger than the mean 15.

In a data set {-21, 5, 8, 10} the mean is (-21+5+8+10)/4=2/4=0.5.
The data set without the outlier of -21 is {5, 8, 10}, the mean is (5+8+10)/4=23/4=7.67.
The mean increased from 0.5 to 7.67, since the outlier -21 is smaller than the mean 0.5.

The effect of an outlier on a median

The median is the middle number in a sorted in ascending order data set.

The median is calculated from the middle value/ values of the data set, therefore removing the outlier will change the positions of the values that are taken to calculate the median.

If we remove a small outlier (the first number), the positions of all values get smaller by 1 (value number 2 becomes number 1, value number 3 becomes number 2…). After removing a small outlier, we take values in bigger positions to calculate the median. Since the values are in ascending order, the numbers in bigger positions have bigger values, therefore the median will be bigger (the median will be same if the values in the positions after the removal are equal to the values in the positions before the removal).   

If we remove a large outlier (the last number), the positions of all values don’t change but the positions of the values that are taken to calculate the median get smaller by 1. Since the values are in ascending order, the numbers in smaller positions have smaller values, therefore the median will be smaller (the median will be same if the values in the positions after the removal are equal to the values in the positions before the removal).

Removing a small outlier will enlarge the median; removing a large outlier will reduce the median (the median will be the same if the values in the positions after the removal are equal to the values in the positions before the removal).

Consider the following example:

Calculate the median of the following data sets with and without the outlier:

{5, 7, 8, 40}

{-21, 5, 8, 10}

In a data set of {5, 7, 8, 40} the median is (7+8)/2=7.5.
The data set without the outlier of 40 is {5, 7, 8}, its median is 7.
The median decreased from 7.5 to 7 after removing a big outlier of 40.

In a data set {-21, 5, 8, 10} the median is (5+8)/2=13/2=6.5.
The data set without the outlier of -21 is {5, 8, 10}, its median is 8.
The median increased from 6.5 to 8 after removing a small outlier of -21.

Note that if the numbers in the new locations (after removing the median) are equal to the numbers in the previous locations (before removing the median), the median will be the same.

The effect of an outlier on the range

The range formula is: range= the highest value – the lowest value.

Since an outlier is a very small or a very big value, it is included in the range calculation, therefore removing the outlier will affect the value of the range. The value that will replace the outlier will be less distant therefore the range after removing the outlier will be smaller.

Consider the following example:

Calculate the range of the following data sets with and without the outlier:

{5, 7, 8, 40}

{-21, 5, 8, 10}

In a data set of {5, 7, 8, 40} the range is 40-5=35.
The data set without the outlier of 40 is {5, 7, 8}, its range is 8-5=3.
The range decreased from 35 to 3 after removing the outlier.

In a data set {-21, 5, 8, 10} the range is 10–21=10+21=31.
The data set without the outlier of -21 is {5, 8, 10}, its range is 10-5=5.
The range decreased from 31 to 5 after removing the outlier.

The effect of an outlier on a standard deviation

The standard deviation is the measure of the overall spread of a data set values from the mean. Since the outlier is a value that is far from other values and the mean, its removal will reduce the spread of the data and the standard deviation.  

Consider the following example:

Calculate the standard deviation of the following data sets with and without the outlier:

{5, 7, 8, 40}

{-21, 5, 8, 10}

In a data set of {5, 7, 8, 40} the mean is 15. The sum of the distances from the mean is |40-15|+|8-15|+|7-15|+|5-15|=25+7+8+10=50.

The data set without the outlier of 40 is {5, 7, 8}, the mean is 6.67. The sum of the distances from the mean is |8-6.67|+|7-6.67|+|5-6.67|=1.33+0.33+1.67=3.33.

The sum of the distances from the mean reduced from 50 to 3.33, therefore the standard deviation of the data set significantly reduced.

In a data set {-21, 5, 8, 10} the mean is 0.5. The sum of the distances from the mean is |-21-0.5|+|5-0.5|+|8-0.5|+|10-0.5|=21.5+4.5+7.5+9.5=43.

The data set without the outlier of -21 is {5, 8, 10}, the mean is 7.67. The sum of the distances from the mean is |5-7.67|+|8-7.67|+|10-7.67|=2.67+0.33+2.33=5.33.

The sum of the distances from the mean reduced from 43 to 5.33, therefore the standard deviation of the data set significantly reduced.

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Data collection and data inference

Data collection and data inference on the SAT test

SAT Subscore: Problem solving and data analysis

Studying data collection and data inference

On the SAT test data collection and data inference topic is part of problem solving and data analysis subscore that includes 9 advanced topics (see the full topics list on the top menu). 

Data collection and data inference topic is the sixth topic of problem solving and data analysis subscore. It is recommended to start learning problem solving and data analysis subscore with its first topic called ratios, rates and proportions.

Data collection and data inference topic is divided into sections from easy to difficult (the list of the sections appears on the left menu). Each section includes detailed explanations of the required material with examples followed by a variety of self-practice questions with solutions.

Finish studying heart of algebra subscore topics before you study this topic or any other problem solving and data analysis subscore topic. (Heart of algebra subscore includes basic algebra topics which knowledge is required for understanding problem solving and data analysis subscore topics). 

Data collection and data inference- summary

Data collection

Data collection is a process of collecting and measuring information on variables of interest, that enables the researcher to test hypotheses and evaluate outcomes.

Data can be collected with a sample or with a controlled experiment:

A sample is a small group that is selected from a large population by using a pre- defined sampling method. The sample must be representative and random.

A controlled experiment is an experiment made on an experimental group, while one factor that is being tested is changed by the researchers and all other factors are held constant.
Each controlled experiment must have a control group. In the control group we don’t change the factor that is being tested in the experimental group. The participants of the control group must be randomly selected and must closely resemble the participants in the experimental group.

Data inference

Data inference is a generalization about a population that is based on statistics calculated from a small group (a sample) that is drawn from that population.

An estimate is a process of finding a value of a population that is close enough to the right value, by performing a sample on a part of that population.
A sample proportion is a variable that is calculated from the sample, that we assume reflects the whole population.
The estimate formula: estimate= sample proportion * population

A margin of error is the degree of error in results received from random sampling surveys, it exists since the sample does not exactly match the population.
The range formula: range= estimate ± margin of error

Continue reading this page for detailed explanations and examples.

Data collection

Data collection is a process of collecting and measuring information on variables of interest, that enables the researcher to test hypotheses and evaluate outcomes. Data can be collected with a sample or with a controlled experiment.

Data collection with a sample

A sample is a small group that is selected from a large population by using a pre- defined sampling method. Samples are used when the population size is too large for the test to include the whole population. The three most common types of sample surveys are e-mail surveys, telephone surveys, and interview surveys.

Sample characteristics:
The sample must be representative and random.
A representative sample is a sample that accurately reflects the examined characteristic of the whole population. A sample that includes members that don’t belong to the population is not representative.
A random sample is a sample that was chosen randomly (purely by chance) so that every member of the population has an equal chance to be selected. A sample that over presents or under presents the subgroup is not random.

Consider the following example:

The school principal wants to estimate how many parents in a group of 80 pupils want their children to participate in an activity. She was suggested five sampling methods, are the methods appropriate?

Method 1: Surveying the parents of randomly chosen 20 girls.
Method 2: Surveying 20 randomly chosen pupils.
Method 3: Surveying 20 randomly chosen parents of the students from student council.
Method 4: Sending the survey with SMS messages to randomly chosen 20 parents of students in the school.
Method 5: Surveying 3 randomly chosen parents of the students.

Method 1: Surveying the parents of a randomly chosen 20 girls. This surveying method is bad because it not chosen randomly. It over presents the sub- group of the girls and under presents the subgroup of the boys.

Method 2: Surveying 20 randomly chosen pupils. This surveying method is bad because it is not representative. It includes pupils that are not a part of the population, since the population is the parents.

Method 3: Surveying 20 randomly chosen parents of the students from student council. This surveying method is bad because it not chosen randomly. It over presents the sub- group of parents of the students from student council.

Method 4: Sending the survey with SMS messages to randomly chosen 20 parents of students in the school. This surveying method is bad because it not chosen randomly. It over presents the sub- group of parents of students that feel positive about the activity and therefore more likely to answer the SMS.

Method 5: Surveying 3 randomly chosen parents of the students. This surveying method is bad because it is not representative. It includes only 3 parents from a group of parents of 80 pupils, therefore it is too small.

 Data collection with a controlled experiment

We can conduct a controlled experiment and conclude about the population from the experiment outcomes.

A controlled experiment is an experiment made on an experimental group, while one factor that is being tested is changed by the researchers and all other factors are held constant (like they were before the experiment).

An independent variable is a variable that is being changed by the researchers in the experimental group.

A correlation means that there is a relationship between two variables (a positive correlation means the variables change at a same direction).

A causation means that a change in one variable is the cause of a change in another variable. In other words: one event is the result of the occurrence of the other event.

Note that a correlation is not a causation. If two variables correlate it does not mean that one causes the other, since the correlation may be caused by a third variable that affects both variables.

For example:
If we want to test the effect of a new medication, the experimental group should include 1,000 participants that will receive the medication for 1 month. The independent variable is taking the new medication for 1 month. All the other factors, like the use of other medications, must remail constant, because changing other factors may influence the outcome of the experiment and we might wrongly assume that the influence came from the factor being tested. (If a participant stops taking the medication for lowering the blood pressure his blood pressure will rise and the researchers can wrongly assume that the rise in the blood pressure was a result of a new medication).

Consider the following example:

Which experiment is appropriate to test the effect of a new quit smoking treatment on the population of smokers?

A. An experiment conducted on 500 smokers aged 20-40 that received the new treatment.

B. An experiment conducted on 500 smokers that received the new treatment and were put on a diet.

C. An experiment conducted on 500 smokers that received the new treatment.

D. An experiment conducted on 500 smokers that had a low blood pressure and received the new treatment.

The answer A in not correct, since the sample is not random, the sample over presents the subgroup of men aged 20-40.

The answer B in not correct, since there are 2 independent variables instead of one. In addition to receiving the new treatment, the smokers were also put on a diet.

The answer D in not correct, since the sample is not random, the sample over presents the subgroup of men that had a low blood pressure.

The answer C in correct, the sample is random and has one independent variable.

The control group in a controlled experiment

In a control group, we don’t change the factor that is being tested in the experimental group (it stays like it was before the experiment). Meaning that all the factors are identical between the two groups except for the factor being tested.

A control group properties:

  • The participants must be randomly selected to be in the control group.
  •  The participants must closely resemble the participants who are in the experimental group.

The purpose of the control group is to rule out alternative explanations of the experimental results.

For example:
What is the control group in the previous example?
Composition: The control group is composed of participants who do not receive the experimental medication. In addition, all the conditions must be unchanged (stay like they were before the experiment). The participants of the control group must resemble the participants of the experimental group in their age, medical conditions and other parameters.
Purpose: An unknown disease may occur in the experimental group and the control group. The existence of the control group will reveal the unknown disease as an unexpected factor, otherwise the researchers will wrongly assume that the unknown disease was caused by the new medicine.

Consider the following example:

Which control group is appropriate to test the effect of a new quit smoking treatment on the population of smokers?

A. A group of 500 smokers that received the new treatment.

B. A group of 500 male smokers.

C. A group of 500 smokers that were put on a diet.

D. A group of 500 smokers.

The answer A in not correct, since in the control group we don’t change the factor that is being tested, so that all the conditions must be unchanged (the smokers in the control group shouldn’t receive the treatment).

The answer B in not correct, since the participants of the control group don’t resemble the participants of the experimental group (there are only male participants in the control group).

The answer C in not correct, since in the control group we don’t change any factor, so that all the conditions must be unchanged (the smokers in the control group shouldn’t be put on a diet).

The answer D in correct, since all the factors are unchanged and the participants of the control group resemble the participants of the experimental group.

Data inference

Data inference is a generalization about a population that is based on statistics calculated from a small group (representative sample) that is drawn from that population. In other words: Instead of checking the whole population we check only a part of the population (representative sample) and assume that the conclusion that was derived from the representative sample is relevant to the whole population.

An estimate calculation

An estimate is a process of finding a value of a population that is close enough to the right value, by performing a sample on a part of that population.

A sample proportion is a variable that is calculated from the sample, that we assume reflects the whole population. A sample proportion can be written as fraction or as a percentage. For example: 10 percent of the sample have a positive opinion about the surveyed subject.

The estimate formula: If we found that a certain percentage from the sample (a sample proportion) represents the percentage in the whole population, we can calculate an estimate by multiplying that percentage by the total amount of items in the population.

estimate= sample proportion * population

Consider the following example:

The school principal wants to estimate the number of pupils that will participate in an activity. She makes a representative sample of 30 pupils and 10 of them answer that they will participate in the activity.

If there are 8 classes in the school with 20 pupils in each class, how many pupils are expected to participate in the activity?

The sample proportion is 10/30=33.33%=0.3333.

The population is 8*20=160 pupils.

estimate= sample proportion * population
estimate= 0.3333 * 160=53 pupils.

The answer is that approximately 53 pupils are expected to participate in the activity.

A range calculation

A margin of error is the degree of error in results received from random sampling surveys, it exists since the sample does not exactly match the population. The margin of error is commonly given as a percentage and is added to the estimate to increase the confidence in the estimate.

Note that:

  • Even after including a margin of error, there is no certainty that the estimation is correct.
  • A high margin of error indicates small confidence that the results represent the population.
  • The larger the sample, the smaller the margin of error (bigger sample size increases the certainty in the prediction).

The range formula: We add (or subtract) the margin of error to the estimate to display the size of the error getting an outcome of a range instead of a single estimate.

range= estimate ± margin of error

Consider the following example:

15 percent of the sample participants own a dog and the margin of error for the sample is 2 percent. If there are 3,000 residents in the town, how many of them are expected to own a dog?

sample proportion= 15%

population= 3,000

estimate= sample proportion * population
estimate= 15% * 3,000= 0.15*3,000=450

the margin of error= 2% from the estimate= 0.02*450=9

The number of dog owners is 450±9, between 441 and 459.

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