Question

D is a point on the side BC of ∆ABC. A line PDQ, through D, meets side AC in P and AB produced at Q. If ∠A = 80°, ∠ABC = 60° and ∠PDC = 15°, find (i) ∠AQD (ii) APD.

Answer 1

$$\begin{gathered} \angle \mathrm{ABD}\mathrm{and}\angle \mathrm{QBD}\mathrm{form}\mathrm{a}\mathrm{linear}\mathrm{pair}. \\ \therefore \angle \mathrm{ABC}+\angle \mathrm{QBC}=180° \\ \Rightarrow 60°+\angle \mathrm{QBC}=180° \\ \mathbf{\angle }\mathbf{QBC}\mathbf{=}\mathbf{120}\mathbf{°} \\ \angle \mathrm{PDC}=\angle \mathrm{BDQ}(\mathrm{Vertically}\mathrm{opposite}\mathrm{angles}) \\ \Rightarrow \angle \mathrm{BDQ}=15° \\ \mathrm{In}△\mathrm{QBD}: \\ \angle \mathrm{QBD}+\angle \mathrm{QDB}+\angle \mathrm{BQD}=180°(\mathrm{Sum}\mathrm{of}\mathrm{angles}\mathrm{of}△\mathrm{QBD}) \\ 120°+15°+\angle \mathrm{BQD}=180° \\ \angle \mathrm{BQD}=180°-135° \\ \angle \mathrm{BQD}=45° \\ \mathbf{\angle }\mathbf{AQD}\mathbf{=}\mathbf{\angle }\mathbf{BQD}\mathbf{=}\mathbf{45}\mathbf{°} \\ \mathrm{In}△\mathrm{AQP}: \\ \angle \mathrm{QAP}+\angle \mathrm{AQP}+\angle \mathrm{APQ}=180°(\mathrm{Sum}\mathrm{of}\mathrm{angles}\mathrm{of}△\mathrm{AQP}) \\ 80°+45°+\angle \mathrm{APQ}=180° \\ \mathbf{\angle }\mathbf{APQ}\mathbf{=}\mathbf{55}\mathbf{°} \\ \mathbf{\angle }\mathbf{APD}\mathbf{=}\mathbf{\angle }\mathbf{APQ} \end{gathered}$$