Question

In Fig. 10.85, PQ is a chord of a circle and PT is the tangent at P such that $\angle$QPT = 60⁰. Then , find $\angle$PRQ .

figure

Answer 1

Construction: Take any point on major arc PQ and name it S. Join SQ and SP.
In the given figure, PT is the tangent. So, PT ⊥ PO.
$\angle \mathrm{QPT}=60°$.
Thus, $\angle \mathrm{OPQ}=90°-60°=30°$
OQ = OP (Radii of the circle)
$\angle \mathrm{OQP}=\angle \mathrm{OPQ}=30°$
$$\begin{gathered} \mathrm{In}△\mathrm{OPQ}, \\ \angle \mathrm{OPQ}+\angle \mathrm{OQP}+\angle \mathrm{POQ}=180° \\ \Rightarrow 30°+30°+\angle \mathrm{POQ}=180° \\ \Rightarrow \angle \mathrm{POQ}=120° \end{gathered}$$
Now, $\angle \mathrm{PSQ}=\frac{1}{2}\angle \mathrm{POQ}=\frac{1}{2}\times 120=60°$
PSQR is a cyclic quadrilateral. Thus,
$$\begin{gathered} \angle \mathrm{PSQ}+\angle \mathrm{PRQ}=180° \\ \Rightarrow 60°+\angle \mathrm{PRQ}=180° \\ \Rightarrow \angle \mathrm{PRQ}=120° \end{gathered}$$