Question

In the given figure, AP and AQ are the tangents drawn to a circle from a point A outside the circle. If $\angle$PAQ = ${40}^{\circ }$ then, find $\angle$AQP.

• A. ${90}^{\circ }$
• B. ${35}^{\circ }$
• C. ${70}^{\circ }$
• D. ${20}^{\circ }$

Answer 1

The correct option is C ${70}^{\circ }$

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

Given : $\angle$PAQ = ${40}^{\circ }$
so, $\angle$OQP = $\frac{\angle PAQ}{2}$
$\Rightarrow$ $\angle$OQP = $\frac{{40}^{\circ }}{2}$
$\Rightarrow$ $\angle$OQP = ${20}^{\circ }$
By Theorem : The tangent at any point of a circle is perpendicular to the radius through the point of contact.
so, $\angle$OQA = ${90}^{\circ }$
As, $\angle$OQA = $\angle$OQP + $\angle$AQP
$\Rightarrow$ ${90}^{\circ }$ = ${20}^{\circ }$ + $\angle$AQP
$\Rightarrow$ $\angle$AQP = ${90}^{\circ }$ - ${20}^{\circ }$
$\Rightarrow$ $\angle$AQP = ${70}^{\circ }$