Question

In the given figure, two lines segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°, then ∠PBA = ?


(a) 50°
(b) 30°
(c) 60°
(d) 100°

Answer 1

(d) 100°

$$\begin{gathered} \mathrm{In}△APB\mathrm{and}△DPC,\mathrm{we}\mathrm{have}: \\ \angle APB=\angle DPC=50° \\ \frac{AP}{BP}=\frac{6}{3}=2 \\ \frac{DP}{CP}=\frac{5}{2.5}=2 \\ \mathrm{Hence},\frac{AP}{BP}=\frac{DP}{CP} \\ \mathrm{Applying}SAS\mathrm{theorem},\mathrm{we}\mathrm{conclude}\mathrm{that}△APB~△DPC. \\ \therefore \angle PBA=\angle PCD \\ \mathrm{In}△DPC,\mathrm{we}\mathrm{have}: \\ \angle CDP+\angle CPD+\angle PCD=180° \\ \Rightarrow \angle PCD=180°-\angle CDP-\angle CPD \\ \Rightarrow \angle PCD=180°-30°-50° \\ \Rightarrow \angle PCD=100° \\ \mathrm{Therefore},\angle PBA=100° \end{gathered}$$