Question

In the given figure, O is the centre of the circle. If A is any point on minor arc BC, then $\angle BAC-\angle OBC=$ _________.

• A. ${45}^o$
• B. ${135}^o$
• C. ${150}^o$
• D. ${90}^o$

Answer 1

Answer: (D)

${90}^o$

$\angle BAC$ and $\angle BOC$ are supplementary.

$\therefore \angle BOC+\angle BAC={180}^o\dots eqn\left(1\right)$

and $\angle OBC=\angle OCB(\because OB=OC)$

$\Rightarrow \angle OBC={90}^o-\angle BOC$

Putting the above value of $\angle BOC$ in equation (1), we get

$\angle {90}^o-\angle OBC+\angle BAC={180}^o$

$\therefore \angle BAC-\angle OBC={90}^o$