Question

In the given figure, PT is a tangent of a circle, with centre O, at point R. If diameter SQ is produced, it meets with PT at point P with $\angle SPR=x\text{and}\angle QSR=y$, then find the value of x +2y (in degrees).
[3 Marks]

Answer 1

$\text{In the given figure,}OR=OS\text{(Radii)}$

$\therefore \angle ORS=\angle OSR=y\text{(Angles opposite to equal sides are}\text{equal)}$ [1 Mark]

$\text{Also,}\angle ORP={90}^{\circ }\text{(Radius of a circle is perpendicular to}\text{the tangent at the point of contact.)}$

$\text{So,}\angle PRS=\angle ORP+\angle ORS={90}^{\circ }+y$ [1 Mark]

$\text{In}\Delta PRS,$
$\angle SPR+\angle PSR+\angle PRS={180}^{\circ }\text{(Angle sum property of a triangle)}$
$\Rightarrow x+y+{90}^{\circ }+y={180}^{\circ }\therefore x+2y={90}^{\circ }$ [1 Mark]