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 ∠SPR=x and ∠QSR=y,then find the value of x+2y (in degrees).
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Solution
In the given figure, OR=OS (Radii)
∴∠ORS=∠OSR=y(Angles opposite to equal sides are equal)
Also, ∠ORP=90∘(Radius of a circle is perpendicular to the tangent at the point of contact.)
So, ∠PRS=∠ORP+∠ORS=90∘+y
In ΔPRS, ∠SPR+∠PSR+∠PRS=180∘(Angle sum property of a triangle) ⇒x+y+90∘+y=180∘∴x+2y=90∘