Question

In the following figure, ∠ACB = 40º. Find ∠OAB.

Answer 1

Given, ∠ACB = 40°

We know that, a segment subtends an angle to the circle is half the angle subtends to the centre.

∴ ∠AOB = 2∠ACB

⇒ `∠ACB = (∠AOB)/2`

⇒ `40^circ = 1/2`∠AOB

⇒ ∠AOB = 80°   ...(i) [Both are the radius of a circle]

In ΔAOB, AO = BO  

⇒ ∠OBA = ∠OAB  ...(ii) [Angles opposite to the equal sides are equal]

We know that, the sum of all three angles in a triangle AOB is 180°.

∴ ∠AOB + ∠OBA + ∠OAB = 180°

⇒ 80° + ∠OAB + ∠OAB = 180°   ...[From equations (i) and (ii)] 

⇒ 2∠OAB = 180° – 80°

⇒ 2∠OAB = 100°

∴ ∠OAB = `100^circ/2` = 50°