Question

In the given figure, AP and AQ are the tangents drawn to a circle from a point A outside the circle. If $\angle$PQA= ${65}^{\circ }$ then, find $\angle$PAQ.

• A. ${100}^{\circ }$
• B. ${25}^{\circ }$
• C. ${90}^{\circ }$
• D. ${50}^{\circ }$

Answer 1

The correct option is D ${50}^{\circ }$

By Theorem : The tangent at any point of a circle is perpendicular to the radius through the point of contact.
so, $\angle$OQA = ${90}^{\circ }$
Given: $\angle$PQA = ${65}^{\circ }$
As, $\angle$OQA = $\angle$OQP + $\angle$PQA
$\Rightarrow$ ${90}^{\circ }$ = $\angle$OQP + ${65}^{\circ }$
$\Rightarrow$ $\angle$OQP = ${90}^{\circ }$ - ${65}^{\circ }$
$\Rightarrow$ $\angle$OQP = ${25}^{\circ }$
By theorem : If two tangents AP and AQ are drawn to a circle with centre O from an external point A, then
$\angle$PAQ= 2$\angle$OPQ= 2$\angle$OQP

so, $\angle$PAQ = 2$\angle$OQP
$\Rightarrow$ $\angle$PAQ = 2 $\times$ ${25}^{\circ }$
$\Rightarrow$ $\angle$PAQ = ${50}^{\circ }$