Question

$\text{In the given figure, O is the centre of the}\text{circle and SR is a tangent. If}\angle SRT={65}^{\circ },\text{then find}\angle y.$

• A. ${90}^{\circ }$
• B. ${65}^{\circ }$
• C. ${55}^{\circ }$
• D. ${50}^{\circ }$

Answer 1

The correct option is D ${50}^{\circ }$

We know that, a tangent at any point of a circle is perpendicular to the radius through the point of contact.

In the given figure, SR is a tangent.
$\text{So, SR}\perp \text{ST.}\therefore \angle TSR={90}^{\circ }$

$\text{Now, consider}△TSR,\text{By angle sum property,}\angle RTS+\angle TSR+\angle SRT={180}^{\circ }$

$\text{[Given:}\angle SRT={65}^{\circ }]\Rightarrow \angle RTS={180}^{\circ }-{65}^{\circ }-{90}^{\circ }\therefore \angle RTS=\angle QTO={25}^{\circ }$

$\text{Consider}△QTO,\text{As,}OT=OQ\text{[Radius of same circle]}$
In a triangle, angles opposite to equal sides are equal.
$\therefore \angle TQO=\angle QTO={25}^{\circ }$

By Exterior angle property,
$\angle SOQ=\angle QTO+\angle TQO\because \angle SOQ=\angle y\therefore \angle y={25}^{\circ }+{25}^{\circ }={50}^{\circ }$