Question

O is the centre of the circle. If $\angle BOA={90}^{\circ }$ and $\angle COA={110}^{\circ }$, find $\angle BAC$.

Answer 1

Given : A circle with centre O and $\angle BOA={90}^{\circ }$ , $\angle COA={110}^{\circ }$. To find : $\angle BAC=?$
$\angle AOB+\angle AOC+\angle BOC={360}^{\circ }$
$\Rightarrow 90°+110°+\angle BOC={360}^{\circ }$
$\Rightarrow \angle BOC={360}^{\circ }–{90}^{\circ }–{110}^{\circ }$
$\Rightarrow \angle BOC={160}^{\circ }$
But, arc BC subtends ∠BOC at the centre and ∠BAC at the remaining part of the circle.
$\Rightarrow \angle BAC=\frac{1}{2}\angle BOC$
$\therefore \angle BAC=\frac{1}{2}\left({160}^{\circ }\right)={80}^{\circ }$