Question

In the given figure, if the angle between two radii of a circle is ${120}^{\circ }$, then find $\angle$PAO.

• A. ${90}^{\circ }$
• B. ${45}^{\circ }$
• C. ${30}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is C ${30}^{\circ }$

Given $\angle$POQ = ${120}^{\circ }$

Consider quadrilateral POQA,

$\angle$POQ + $\angle$OQA + $\angle$QAP + $\angle$APO = ${360}^{\circ }$ -------(1)

[By Theorem - The tangent at any point of a circle is perpendicular to the radius through the point of contact]

So, $\angle$OQA = $\angle$APO = ${90}^{\circ }$

On substituting these values in (1) we get,

$\Rightarrow$ ${120}^{\circ }$ + ${90}^{\circ }$ + $\angle$QAP + ${90}^{\circ }$ = ${360}^{\circ }$

$\Rightarrow$ $\angle$QAP = ${360}^{\circ }$ - ${120}^{\circ }$ - ${90}^{\circ }$ + ${90}^{\circ }$

$\Rightarrow$ $\angle$QAP = ${60}^{\circ }$

Draw a line joining point A and O.

[By Theorem - The centre of a circle lies on the bisector of the angle between two tangents drawn

from a point outside it]

so, line AO bisects $\angle$QAP

$\therefore$ $\angle$PAO = $\angle$QAO

So, $\angle$QAP = $\angle$PAO + $\angle$QAO

$\Rightarrow$ $\angle$QAP = 2 $\times$ $\angle$PAO

$\Rightarrow$ ${60}^{\circ }$ = 2 $\times$ $\angle$PAO

$\Rightarrow$ $\angle$PAO = $\frac{{60}^{\circ }}{2}$

$\Rightarrow$ $\angle$PAO = ${30}^{\circ }$