Question

In fig 2.25, sides QP and RQ of $∆$PQR are produced to the points S and T respectively. If $\angle$SPR = 105$°$, $\angle$PQT = 110$°$ then find $\angle$PRQ.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},\angle \mathrm{SPR}+\angle \mathrm{RPQ}=180°(\mathrm{Linear}\mathrm{pair}) \\ \Rightarrow 105°+\angle \mathrm{RPQ}=180° \\ \Rightarrow \angle \mathrm{RPQ}=180°-105° \\ \Rightarrow \angle \mathrm{RPQ}=75°...\left(\mathrm{i}\right) \\ \mathrm{Also},\angle \mathrm{PQT}+\angle \mathrm{PQR}=180°(\mathrm{Linear}\mathrm{pair}) \\ \Rightarrow 110°+\angle \mathrm{PQR}=180° \\ \Rightarrow \angle \mathrm{PQR}=180°-110° \\ \Rightarrow \angle \mathrm{PQR}=70°....\left(\mathrm{ii}\right) \\ \mathrm{In}△\mathrm{PQR},\mathrm{we}\mathrm{have}, \\ \angle \mathrm{RPQ}+\angle \mathrm{PQR}+\angle \mathrm{PRQ}=180°(\mathrm{Sum}\mathrm{of}\mathrm{the}\mathrm{measure}\mathrm{of}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}) \\ \Rightarrow 75°+70°+\angle \mathrm{PRQ}=180°[\mathrm{From}(\mathrm{i})\mathrm{and}(\mathrm{ii}\left)\right] \\ \Rightarrow 145°+\angle \mathrm{PRQ}=180° \\ \Rightarrow \angle \mathrm{PRQ}=180°-145° \\ \Rightarrow \angle \mathrm{PRQ}=35° \end{gathered}$$