Question:
In the given figure, ∠AOB = 90° and ∠ABC = 30°. Then, ∠CAO = ?
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Solution:
(c) 60°
We have:
∠AOB = 2∠ACB
$\Rightarrow \angle \mathrm{ACB}=\frac{1}{2} \angle \mathrm{AOB}=\left(\frac{1}{2} \times 90^{\circ}\right)=45^{\circ}$
∠COA = 2∠CBA = (2 × 30°) = 60°
∴ ∠COD = 180° - ∠COA = (180° - 60°) = 120°
$\Rightarrow \angle \mathrm{CAO}=\frac{1}{2} \angle \mathrm{COD}=\left(\frac{1}{2} \times 120^{\circ}\right)=60^{\circ}$