Question

In the given figure, AB and AC are tangents to the circle centre O such that $\angle$ BAC = ${40}^o$. Then,$\angle$ BOC is equal to

(a) ${80}^o$ (b) ${100}^o$ (c) ${120}^o$ (d) ${140}^o$

Answer 1

∠ BOC+<BAC=180
∠ BOC=180-40
∠ BOC=140