Choose the correct answer from the given four options:
In below figure, two line segments $ \mathrm{AC} $ and $ \mathrm{BD} $ intersect each other at the point $ \mathrm{P} $ such that $ \mathrm{PA}=6 \mathrm{~cm}, \mathrm{~PB}=3 \mathrm{~cm}, \mathrm{PC}=2.5 \mathrm{~cm}, \mathrm{PD}=5 \mathrm{~cm}, \angle \mathrm{APB}=50^{\circ} $ and $ \angle \mathrm{CDP}=30^{\circ} $. Then, $ \angle \mathrm{PBA} $ is equal to

(A) $ 50^{\circ} $
(B) $ 30^{\circ} $
(C) $ 60^{\circ} $
(D) $ 100^{\circ} $"

Given:

Two line segments \( \mathrm{AC} \) and \( \mathrm{BD} \) intersect each other at the point \( \mathrm{P} \) such that \( \mathrm{PA}=6 \mathrm{~cm}, \mathrm{~PB}=3 \mathrm{~cm}, \mathrm{PC}=2.5 \mathrm{~cm}, \mathrm{PD}=5 \mathrm{~cm}, \angle \mathrm{APB}=50^{\circ} \) and \( \angle \mathrm{CDP}=30^{\circ} \).

To do:

We have to find \( \angle \mathrm{PBA} \).

Solution:

In $\triangle A P B$ and $\triangle C P D$,

$\angle A P B=\angle C P D=50^{\circ}$          (Vertically opposite angles)

$\frac{\mathrm{AP}}{\mathrm{PD}}=\frac{6}{5}$..........(i)

$\frac{\mathrm{BP}}{\mathrm{CP}}=\frac{3}{2.5}$

$\frac{\mathrm{BP}}{\mathrm{CP}}=\frac{6}{5}$.......(ii)

From (i) and (ii), we get,

$\frac{AP}{PD}=\frac{BP}{CP}$

Therefore, by SAS similarity,

$\triangle \mathrm{APB} \sim \triangle \mathrm{DPC}$

This implies,

$\angle A=\angle D=30^{\circ}$          (Corresponding angles of similar triangles)

The sum of the angles of a triangle is $180^{\circ}$

In $\triangle A P B$,

$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{APB}=180^{\circ}$

$30^{\circ}+\angle B+50^{\circ}=180^{\circ}$

$\angle B=180^{\circ}-(50^{\circ}+30^{\circ})$

$\angle B=180-80^{\circ}$

$\angle B=100^{\circ}$

Therefore, $\angle \mathrm{PBA}=100^{\circ}$.