Question

In Figure 2, AB and AC are tangents to the circle with centre O such that $\angle BAC={40}^{\circ }$. Then $\angle BOC$ is equal to

• A. ${40}^{\circ }$
• B. ${50}^{\circ }$
• C. ${140}^{\circ }$
• D. ${150}^{\circ }$

Answer 1

The correct option is C ${140}^{\circ }$

Tangent is perpendicular to radius at point of contact.

So, $\angle ABO=\angle ACO={90}^{\circ }$

In a quadrilateral, the sum of the angles is ${360}^{\circ }$​.

$\angle BAC+\angle BOC+\angle ABO+\angle ACO={360}^{\circ }$

$\therefore \angle BAC+\angle BOC=180^{\circ }$

$\angle BOC={180}^{\circ }-{40}^{\circ }$

$\angle BOC={140}^{\circ }$

So, option C is the answer