Question 8
SSC-CGL 2020 Mar 4 Shift 2
PRT is a tangent to a circle with center O touching the circle at R. Diameter SQ is produced to meet P. If angle PRQ = 28 degree, then find angle SPR.
Solution in Detail
by exterior angle of $\displaystyle \Delta {PQR}$,
$\displaystyle \angle {OQR} = x + 28$
Observe $\displaystyle \Delta {OQR}$ is isosceles:
$\displaystyle \therefore \angle{ORQ} = \angle {OQR} = x + 28$
But radius $\displaystyle OR \perp \text{tangent PRT} $
$\displaystyle \therefore \angle{ORQ} + \angle{QRP} = 90$
$\displaystyle \therefore (x + 28) + 28 = 90$
$\displaystyle \implies x = 34\degree \:\underline{Ans}$
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