Question

In the given figure, O is the centre of a circle and PT is the tangent to the circle. If PQ is a chord, such that ∠QPT = 50° then ∠POQ = ?


(a) 100°
(b) 90°
(c) 80°
(d) 75°

Answer 1

$$\begin{gathered} \left(\mathrm{a}\right)100° \\ \mathrm{Given},\angle QPT={50}^0 \\ \mathrm{and}\angle OPT={90}^0\left(\text{T}\mathrm{angents}\mathrm{}\mathrm{drawn}\mathrm{}\mathrm{from}\mathrm{}\mathrm{an}\mathrm{}\mathrm{external}\mathrm{}\mathrm{point}\mathrm{}\mathrm{are}\mathrm{}\mathrm{perpendicular} \\ \mathrm{to}\mathrm{}\mathrm{the}\mathrm{}\mathrm{radius}\mathrm{}\mathrm{at}\mathrm{}\mathrm{the}\mathrm{}\mathrm{point}\mathrm{}\mathrm{of}\mathrm{}\mathrm{contact}\right) \\ \therefore \angle OPQ=\left(\angle OPT-\angle QPT\right)=\left({90}^0-{50}^0\right)={40}^0 \\ OP=OQ\left(\text{R}\mathrm{adius}\mathrm{}\mathrm{of}\mathrm{}\mathrm{the}\mathrm{}\mathrm{same}\mathrm{}\mathrm{circle}\right) \\ \Rightarrow \angle OQP=\angle OPQ={40}^0 \\ \mathrm{In}∆POQ,\angle POQ+\angle OQP+\angle OPQ={180}^0 \\ \therefore \angle POQ={180}^0-\left({40}^0+{40}^0\right)={100}^0 \end{gathered}$$