Question

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

• A. ${30}^{\circ }$
• B. ${120}^{\circ }$
• C. ${90}^{\circ }$
• D. ${15}^{\circ }$

Answer 1

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

Join $OB$.

We know that the radius and tangent are perpendicular at their point of contact.

$\therefore$ $\angle OBP=\angle OAP={90}^o$

Now, In a quadrilateral $AOBP$

$\Rightarrow$ $\angle AOB+\angle OBP+\angle APB+\angle OAP={360}^o$ [ Sum of four angles of a quadrilateral is ${360}^o.$ ]

$\Rightarrow$ $\angle AOB+{90}^o+{60}^o+{90}^o={360}^o$

$\Rightarrow$ ${240}^o+\angle AOB={360}^o$

$\Rightarrow$ $\angle AOB={120}^o$.

Since $OA$ and $OB$ are the radius of a circle then, $△AOB$ is an isosceles triangle.

$\Rightarrow$ $\angle AOB+\angle OAB+\angle OBA={180}^o$

$\Rightarrow$ ${120}^o+2\angle OAB={180}^o$ [ Since, $\angle OAB=\angle OBA$ ]

$\Rightarrow$ $2\angle OAB={60}^o$

$\therefore$ $\angle OAB={30}^o$