Question

In the given figure, O is the centre of a circle. If $\angle OAB={40}^{\circ }$ and C is a point on the circle then $\angle ACB=?$

(a) ${40}^{\circ }$

(b) ${50}^{\circ }$

(c) ${80}^{\circ }$

(d) ${100}^{\circ }$

Answer 1

(b) 50°
OA = OB
⇒ ∠OBA = ∠OAB = 40°
Now, ∠AOB = 180° - (40° + 40°) = 100°
∴ ∠ACB=1/2∠AOB=1/2×100°=50°