Question

In the given figure, O is the centre of a circle. BOA is its diameter and the tangent at the point P meets BA extended at T. If ∠PBO = 30°, then ∠PTA = ?


(a) 60°
(b) 30°
(c) 15°
(d) 45°

Answer 1

(b) 30°
$$\begin{gathered} \angle BPA={90}^0\left[\because \mathrm{angle}\mathrm{}\mathrm{in}\mathrm{}\mathrm{a}\mathrm{}\mathrm{semicircle}\right] \\ \mathrm{In}△PBA,\angle BPA+\angle PBA+\angle BAP={180}^0 \\ \Rightarrow {90}^0+{30}^0+\angle BAP={180}^0 \\ \Rightarrow \angle BAP=\left({180}^0-{120}^0\right) \\ \Rightarrow \angle BAP={60}^0 \\ BAT\mathrm{is}\mathrm{}\mathrm{a}\mathrm{}\mathrm{straight}\mathrm{}\mathrm{angle}. \\ \therefore \angle BAP+\angle PAT={180}^0 \\ \Rightarrow {60}^0+\angle PAT={180}^0 \\ \Rightarrow \angle PAT=\left({180}^0-{60}^0\right) \\ \Rightarrow \angle PAT={120}^0 \\ OA=OP\left(\text{R}\mathrm{adius}\mathrm{}\mathrm{of}\mathrm{}\mathrm{the}\mathrm{}\mathrm{same}\mathrm{}\mathrm{circle}\right) \\ \Rightarrow \angle OAP=\angle OPA={60}^0\left(S\mathrm{ince}\angle BAP={60}^0\right) \\ \mathrm{N}\mathrm{o}\mathrm{w},\mathrm{}\angle OPT={90}^0(\text{T}\mathrm{angents}\mathrm{}\mathrm{drawn}\mathrm{}\mathrm{from}\mathrm{}\mathrm{an}\mathrm{}\mathrm{external}\mathrm{}\mathrm{point}\mathrm{}\mathrm{are}\mathrm{}\mathrm{perpendicular}\mathrm{} \\ \mathrm{to}\mathrm{}\mathrm{the}\mathrm{}\mathrm{radius}\mathrm{}\mathrm{at}\mathrm{}\mathrm{the}\mathrm{}\mathrm{point}\mathrm{}\mathrm{of}\mathrm{}\mathrm{contact}) \\ \Rightarrow \angle OPA+\angle APT={90}^0 \\ \Rightarrow {60}^0+\angle APT={90}^0 \\ \Rightarrow \angle APT=\left({90}^0-{60}^0\right) \\ \Rightarrow \angle APT={30}^0 \\ In△PAT,\text{we}\mathrm{have}: \\ \angle PAT+\angle APT+\angle PTA={180}^0 \\ \Rightarrow {120}^0+{30}^0+\angle PTA={180}^0 \\ \Rightarrow \angle PTA=\left({180}^0-{150}^0\right) \\ \Rightarrow \angle PTA={30}^0 \end{gathered}$$