Question

In the given figure, AOB is a diameter of a circle and CD||AB. If $\angle BAD={30}^{\circ }then\angle CAD=?$

(a) ${30}^{\circ }$

(b) ${60}^{\circ }$

(c) ${45}^{\circ }$

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

Answer 1

(a) 30°
$\angle ADC$ = $\angle BAD$ = 30° (Alternate angles)
$\angle ADB$ = 90° (Angle in semicircle)
∴ $\angle CBD$ = (90° + 30°) = 120°
But ABCD being a cyclic quadrilateral, we have:
$\angle BAC$ + $\angle CDB$ = 180°
⇒ $\angle BAD$ + $\angle CAD$ + $\angle CDB$ = 180°
⇒ 30° + $\angle CAD$ + 120° = 180°
⇒ $\angle CAD$ = (180° - 150°) = 30°