# Angles, arc lengths and trig functions on the SAT test

### SAT Subscore: Additional topics in math

**Angles, arc length and trig functions subject includes 3 topics: **

Calculations of angles in radians.

Calculation of arc lengths and sector areas in radians.

Calculation of sine, cosine and tangent in radians.

**Before learning this subject learn the pages about circle theorems and right triangle trigonometry.**

**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).

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=πr^{2}.

A=π*10^{2}

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.

#### 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).

#### 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?

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).

**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°=2____π____ 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.