Question

Two tangents are drawn from an external point P (as shown in fig.) such that $\angle OBA={10}^{\circ }$ . Then angle $\angle BPA$ is.

• A. ${10}^{\circ }$
• B. ${20}^{\circ }$
• C. ${30}^{\circ }$
• D. ${40}^{\circ }$

Answer 1

The correct option is B ${20}^{\circ }$

Since tangent and radius are perpendicular at the point of contact, we have:

$\angle OAP=\angle OBP={90}^{\circ }$.

So, $\angle AOB+\angle BPA={180}^{\circ }\dots \left(1\right)$

In $△OAB$, $OA=OB$, so, $\angle OAB=\angle OBA={10}^{\circ }$.

So, $\angle AOB={180}^{\circ }-{10}^{\circ }-{10}^{\circ }={160}^{\circ }\dots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, we get $\angle BPA={180}^{\circ }-{160}^{\circ }={20}^o$.

So option B is the right answer.