Question

In the given figure, $△ABC$ is inscribed in a circle. The bisector of $\angle BAC$ meets BC at D and the circle at E. If EC is joined, then $\angle ECD={30}^{\circ }$. Find $\angle BAC$.

• A. 30°
• B. 60°
• C. 50°
• D. 70°

Answer 1

The correct option is B

60°

$\because$ Angles in same segment are equal

$\angle BCE$ = $\angle BAE$ = ${30}^{\circ }$

$\because$ AD is a bisector

So $\angle BAE$ = $\angle CAE$ = ${30}^{\circ }$

$\therefore$ $\angle BAC$ =${30}^{\circ }$ + ${30}^{\circ }$ = ${60}^{\circ }$