Question

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

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

Answer 1

The correct option is A

${90}^{\circ }$

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

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

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

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