Question

In figure, $O$ is the centre of the circle, $CA$ is tangent at $A$ and $CB$ is tangent at $B$ drawn to the circle. If $\angle ACB={75}^o$, $\angle AOB=$

• A. ${75}^o$
• B. ${85}^o$
• C. ${95}^o$
• D. ${105}^o$

Answer 1

The correct option is D ${105}^o$

Given- $O$ is the centre of a circle to which two tangents, $CA\&CB$ have been drawn at $A\&B$ respectively. $\angle ACB={75}^o$. To find out- $\angle AOB=?$ Solution- $OA\&OB$ are radii drawn from $O$ to $A\&B$ respectively.
$\therefore \angle OBC={90}^o=\angle OAC$ since the radius through the point of contact of a tangent to a circle is perpendicular to the tangent. Now, considering the quadrilateral $AOBC$, we have $\angle OBC+\angle OAC+\angle ACB+\angle AOB={360}^o$ (by angle sum property of quadrilateral)
$⟹{90}^o+{90}^o+{75}^o+\angle AOB={360}^o⟹\angle AOB={105}^o.$
Ans- Option D.