Question

In figure 2, PA and PB are tangents to the circle with centre O. If $\angle$APB = ${60}^{\circ }$, then $\angle$OAB is

• A. 30°
• B. 60°
• C. 90°
• D. 15°

Answer 1

The correct option is A

30°

Since PA and PB are tangents to the circle from an external point O.

Therefore, PA = PB
$\therefore \Delta PAB$ is an isosceles triangle where $\angle PAB=\angle PBA$
$\angle P+\angle PAB+\angle PBA={180}^{\circ }$ [angle sum property of triangle]
$\Rightarrow {60}^{\circ }+2\angle PAB={180}^{\circ }$
$\Rightarrow 2\angle PAB={180}^{\circ }-{60}^{\circ }={120}^{\circ }$
$\Rightarrow \angle PAB=\frac{120}{2}={60}^{\circ }$

It is known that the radius is perpendicular to the tangent at the point of contact.
$\therefore \angle OAP={90}^{\circ }$
$\Rightarrow \angle PAB+\angle OAB={90}^{\circ }$
$\Rightarrow \angle OAB={90}^{\circ }-{60}^{\circ }={30}^{\circ }$
The correct answer is A