Question

In the given figure, AOB is a diameter and ABCD is a cyclic quadrilateral. If $\angle ADC={120}^{\circ }\text{then}\angle BAC=?$

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

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

(c) ${20}^{\circ }$

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

Answer 1

ANSWER:
( b ) 30°
We have:
$\angle ABC$ + $\angle ADC$ = 180° (Opposite angles of a cyclic quadrilateral)
⇒ $\angle ABC$ + 120 ° = 180°
⇒ $\angle ABC$ = (180° - 120°) = 60°
Also, $\angle ACB$ = 90° (Angle in a semicircle)
In ΔABC, we have:
$\angle BAC$ + $\angle ACB$ + $\angle ABC$ = 180° (Angle sum property of a triangle)
⇒ $\angle BAC$ + 90° + 60° = 180°
⇒ $\angle BAC$ = (180° - 150°) = 30°