Question

In the given figure, XY and X'Y' are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersects XY at A and X'Y' at B. Then $\angle$ AOB is _____.

• A. ${80}^{\circ }$
• B. ${60}^{\circ }$
• C. ${85}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is D ${90}^{\circ }$

In quadrilateral APQB, we have

$\angle APO={90}^{\circ }\text{and}\angle BQO={90}^{\circ }$
[$\because$ Tangent is perpendicular to the radius at the point of contact- By Theorem].

Now, $\angle APO+\angle BQO+\angle QBC+\angle PAC={360}^{\circ }$
$\Rightarrow \angle PAC+\angle QBC={360}^{\circ }-(\angle APO+\angle BQO)={180}^{\circ }$ . . . (i)

We have

$\angle CAO=\frac{1}{2}\angle PAC\text{and}\angle CBO=\frac{1}{2}\angle QBC$ [$\because$ Tangents from an external point are equally inclined to the line segment joining the centre to that point- By Theorem].

On adding $\angle CAO$ and $\angle CBO$ and using (i) we get,

$\angle CAO+\angle CBO=\frac{1}{2}(\angle PAC+\angle QBC)=\frac{1}{2}\times {180}^{\circ }={90}^{\circ }$ . . .(ii)

In $\Delta$ AOB, we have

$\angle CAO+\angle AOB+\angle CBO={180}^{\circ }$
$\Rightarrow \angle AOB={180}^{\circ }-(\angle CAO+\angle CBO)={90}^{\circ }$ [using (ii)].

Hence, $\angle AOB={90}^{\circ }$