Question

In the given figure, O is the centre of a circle, PQ is a chord and the tangent PT at P makes an angle of 50° with PQ. Then, ∠POQ = ?


(a) 130°
(b) 100°
(c) 90°
(d) 75°

Answer 1

$$\begin{gathered} \left(b\right)100° \\ \mathrm{Given},\mathit{\angle QPT}={50}^0 \\ \mathrm{Now},\mathit{\angle OPT}={90}^0(\text{S}\mathrm{ince}\mathrm{tangents}\mathrm{drawn}\mathrm{from}\mathrm{an}\mathrm{external}\mathrm{point}\mathrm{a}\mathrm{re} \\ \mathrm{perpendicular}\mathrm{to}t\mathrm{he}\mathrm{radius}\mathrm{at}\mathrm{point}\mathrm{of}\mathrm{contact}\mathit{)} \\ \mathit{\therefore }\mathit{\angle OPQ}\mathit{=}\left(\mathit{\angle OPT}\mathit{-}\mathit{\angle QPT}\right)\mathit{=}\left({\mathit{90}}^{\mathit{0}}\mathit{-}{\mathit{50}}^{\mathit{0}}\right)\mathit{=}{\mathit{40}}^{\mathit{0}} \\ \mathit{OP}\mathit{=}\mathit{OQ}\left(\mathit{Radii}\mathit{}\mathit{of}\mathit{}\mathit{the}\mathit{}\mathit{same}\mathit{}c\mathit{ircle}\right) \\ \mathit{\Rightarrow }\mathit{\angle OPQ}\mathit{=}\mathit{\angle OQP}\mathit{=}{\mathit{40}}^{\mathit{0}} \\ \mathrm{In}△\mathit{POQ}\mathit{,} \\ \angle \mathit{POQ}\mathit{+}\angle \mathit{OPQ}\mathit{+}\mathit{\angle OQP}\mathit{=}{\mathit{180}}^{\mathit{0}} \\ \mathit{\Rightarrow }\angle \mathit{POQ}\mathit{+}{\mathit{40}}^{\mathit{0}}\mathit{+}{\mathit{40}}^{\mathit{0}}\mathit{=}{\mathit{180}}^{\mathit{0}} \\ \mathit{\Rightarrow }\angle \mathit{POQ}\mathit{=}{\mathit{180}}^{\mathit{0}}\mathit{-}\left({\mathit{40}}^{\mathit{0}}\mathit{+}{\mathit{40}}^{\mathit{0}}\right) \\ \mathit{\Rightarrow }\mathit{\angle POQ}\mathit{=}{\mathit{180}}^{\mathit{0}}\mathit{-}{\mathit{80}}^{\mathit{0}} \\ \mathit{\Rightarrow }\mathit{\angle POQ}\mathit{=}{\mathit{100}}^{\mathit{0}} \end{gathered}$$