Question

In the given figure, O is the centre of a circle, ∠AOB = 90° and ∠ABC = 30°. Then, ∠CAO = ?
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Answer 1

(c) 60°

OA = OB
⇒∠OAB = ∠OBA = 45°
$\angle \mathrm{ACB}=\frac{1}{2}\angle \mathrm{AOB}=\left(\frac{1}{2}\times 90°\right)=45°$
In Δ ABC, we have:
∠ACB + ∠CAB + ∠ABC = 180°
⇒ 45° + ∠CAB + 30° = 180°
⇒ ∠CAB = (180° - 75°) = 105°
Now, ∠CAO + ∠OAB = 105°
⇒ ∠CAO + 45° = 105°
⇒ ∠CAO = (105° - 45°) = 60°