Angles created by intersection of diameters in a circle example

Example- angles created by intersection of diameters in a circle
Angles created by intersection of diameters- example

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 (mark them as 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 (mark them as 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=3.5

Leave a comment