Question

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

• A. ${45}^{\circ }$
• B. ${60}^{\circ }$
• C. ${90}^{\circ }$
• D. ${150}^{\circ }$

Answer 1

The correct option is C

${90}^{\circ }$

$\text{In the given figure,}OR=OS\text{(Radii)}$
$\therefore \angle ORS=\angle OSR=y\text{(Angles opposite to equal sides are}\text{equal)}$

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

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

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