Question

In figure, A, B and C are three points on a circle with

Centre O such that ∠BOC = 30° and ∠AOB = 50°. If D

is a point on the circle other than the arc ABC. Find ∠ADC.​

Answer 1

$\bf \large\underline{Solution :- }$

$\sf{AOC=AOB+BOC}$

$\sf{So,\:AOC=60°+30°}$

$\sf{→AOC = 90°}$

• It is known that an angle which is subtended by an arc at the centre of the circle is double the angle subtended by that arc at any point on the remaining part of the circle.

$\sf{So,\:ADC=(\frac{1}{2})AOC}$

$\sf{→ADC=\frac{1}{2}×90°}$

$\sf{→ADC=45°}$

Answer 2

Answer:

According to theorem of circles ∠AOC=2∠ADC ...(i)

∠AOC=∠AOB+∠BOC=60° +30°=90°

from equation(i)

90°=2∠ADC

∠ADC=45

Thanks!!