Question

In the figure shown, if $PA$ and $PB$ are tangents to the circle with centre $O$ such that $\angle P={50}^0$, then $\angle OAB$ is equal to

• A. ${25}^0$
• B. ${30}^0$
• C. ${40}^0$
• D. ${50}^0$

Answer 1

The correct option is A ${25}^0$

Given-

$PA$ & $PB$ are tangents, to the circle with $O$ as centre, at $A$ & $B$ respectively.

$\angle APB={50}^0$

To find:

$\angle OAB=?$

Solution:

In $\Delta OAB$, we have $OA=OB$ (radii of the same circle)

$\therefore \Delta OAB$ is isosceles

$\Rightarrow \angle OAB=\angle OBA$

$\therefore \angle OAB+\angle OBA=2\angle OAB$

$\therefore 2\angle OAB+\angle AOB={180}^0$

$\Rightarrow \angle OAB=\frac{1}{2}({180}^0-\angle AOB)$ (1)

Now $PA$ & $PB$ are tangents to the circle.

$\therefore \angle PAO={90}^0=\angle PBO$

$\Rightarrow \angle PAO+\angle PBO={180}^0$

So, in the quadrilateral $PAOB$ we have

$\angle AOB+\angle APB={360}^0-(\angle PAO+\angle PBO)={180}^0$

(angle sum property of quadrilaterals).

From (1) we have

$\angle OAB=\frac{1}{2}({180}^0-\angle AOB)=\angle OAB=\frac{1}{2}({180}^0-{130}^0)={25}^0$