Question

Two tangents AP and AQ are drawn from an external point A to a circle with centre 'O'. If tangents AP and AQ are inclined to each other at an angle of ${60}^{\circ }$ then, $\angle$POA is ____.

• A. ${120}^{\circ }$
• B. ${60}^{\circ }$
• C. ${70}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is B ${60}^{\circ }$

[By Theorem- The centre of the circle lies on the bisector of the angle between two tangents]
so, line AO bisects $\angle$PAQ
$\therefore$ $\angle$ PAO = $\frac{\angle PAQ}{2}$ = $\frac{{60}^{\circ }}{2}$ = ${30}^{\circ }$

[By Theorem - The tangent at any point of a circle is
perpendicular to the radius through the point of contact]
so, $\angle$ APO = ${90}^{\circ }$

Consider $△$ PAO,
$\angle$ APO + $\angle$ POA + $\angle$ PAO = ${180}^{\circ }$
$\Rightarrow$ ${90}^{\circ }$ + $\angle$ POA + ${30}^{\circ }$ = ${180}^{\circ }$
$\Rightarrow$ $\angle$ POA = ${180}^{\circ }$ - ${90}^{\circ }$ - ${30}^{\circ }$
$\Rightarrow$ $\angle$ POA = ${60}^{\circ }$