Question

In the adjoining figure, $O$ is the center of the circle and $\angle ADB={20}^{\circ }$. Find $\angle BAC$

Answer 1

Given: $\angle ADB={20}^o$

$\Rightarrow \angle ACB={20}^o$

$(\angle ACB=\angle ADB$ as both are angles subtended by same arc $AB$)

& $AC$ passes from centre of the circle

$\Rightarrow \angle ABC={90}^o$

& $\angle ACB={20}^o$ (from above)

$\Rightarrow \angle BAC={180}^o-\angle ABC-\angle ACB$

$=180-90-20$

$={70}^o$.