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

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