﻿ Example of simple circle1 | mathtestpreparation.com back to math video class

### Example of simple circle problem solving1

A simple circle problem solving
In the figure below, point O is the center of the circle. If the angle AOB = 120o, what is the value of x? Solution:
since points A and B are on the circle
so OA = OB = r (r is the radius of the circle.)
since OA = OB, so triangle OAB is an isosceles triangle. In an isosceles triangle, two base angles are equal.
angle OAB = angle OBA
The sum of the interior angles of a triangle is 180o
angle O + angle OAB + angle OBA = 180o
since angle OAB = angle OBA
angle O + 2 angle OAB = 180o
2 angle OAB = 180o - angle O = 180o - 120o = 60o
angle OAB = 60o/2 = 30o
from the figure, angle A = xo
so x = 30
therefore, the value of x is 30.

The distance of any points on the circle to the center of the circle is the radius of the circle. Since point A and point B are on the circle, and the center of the circle is O, so the distance OA is equal to the distance OB. In triangle OAB, since two sides of the triangle are equal, then the triangle is an isosceles triangle. In an isosceles triangle, two base angles are equal. In other word, equal sides of a triangle opposite the equal angles. So the angle A is equal to angle B. In any triangle, the sum of interior angles is 180o. Since angle A + angle B + angle O = 180o. Given angle O = 120o, then angle A + angle B = 180o - 120o = 60o. since angle A = angle B, so two times of the angle A is 60o, so angle A = 30o. Given angle A = xo, so the value of x is 30.