Question

In the adjoining figure, $O$ is the centre of the circle. If $\angle$ $BOA$ $={90}^o$, $\angle$ $COA$ $={128}^o$, find $\angle$ $BAC$.

Answer 1

We have $\angle$ $BOC$ $=\angle$ $AOB$ $+\angle$ $AOC$, and the obtuse angle $\angle$ $BOC$ $={360}^o-({90}^o+{128}^o)={360}^o-{218}^o={142}^o$.

By using "Subtended angle theorem" i.e. [Angle subtended by an arc at the centre is double of the angle subtended by the same arc on the circle]

We have, $\angle BAC=\frac{1}{2}\angle BOC$

$=\frac{1}{2}\left({142}^o\right)={71}^o$.