Question

In the given figure, O is the centre of a circle. If $\angle AOB={100}^{\circ }and\angle AOC={90}^{\circ }then\angle BAC=?$

(a) ${85}^{\circ }$

(b) ${80}^{\circ }$

(c) ${95}^{\circ }$

(d) ${75}^{\circ }$

Answer 1

(a) 85°

We have:
∠BOC + ∠BOA + ∠AOC = 360°
⇒ ∠BOC + 100° + 90° = 360°
⇒ ∠BOC = (360° - 190°) = 170°
∴ ∠BAC=1/2×∠BOC=1/2×170°=85°