Question

In the given figure, equilateral $△$ABC is inscribed in a circle and ABCD is a quadrilateral, as shown. Then $\angle$BDC = ?

(a) ${90}^{\circ }$

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

(c) ${120}^{\circ }$

(d) ${150}^{\circ }$

Answer 1

ANSWER:
( c ) 120°
Since ΔABC is an equilateral triangle, each of its angle is 60°.
∴ $\angle BAC$ = 60°
In a cyclic quadrilateral ABCD, we have:
$\angle BAC$ + $\angle BDC$ = 180°
⇒ 60° + $\angle BDC$ = 180°
⇒ $\angle BDC$ = (180° - 60°) = 120°