Question

Question 5
In figure, two line segments AC and BD intersect each other at the point P such that $PA=6cm,PB=6cm,PC=3cm,PC=2.5cm,PD=5cm,\angle APB={50}^{\circ }and\angle CDP={30}^{\circ }.Then\angle PBA$ is equal to




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

Answer 1

(D) ${100}^{\circ }$

In $\Delta APBand\Delta CPD,$
$\angle APB=\angle CPD={50}^{\circ }$ [ Vertically opposite angles ]
$\frac{AP}{PD}=\frac{6}{5}$............(i)
$\frac{BP}{CP}=\frac{3}{2.5}=\frac{6}{5}$.............(ii)
From Eq.s (i) and (i)
$\frac{AP}{PD}=\frac{BP}{CP}$
$\therefore \Delta APB\sim \Delta DPC$ [ by SAS similarity criterion ]
$\therefore \angle A=\angle D={30}^{\circ }$ [ corresponding angles of similar triangles ]
In $\Delta APB$ ,
$\angle A+\angle B+\angle APB={180}^{\circ }$ [ sum of angles of a triangle = ${180}^{\circ }$ ]
$\Rightarrow {30}^{\circ }+\angle B+{50}^{\circ }={180}^{\circ }$
$\therefore \angle B={180}^{\circ }-({50}^{\circ }+{30}^{\circ })={100}^{\circ }$
i.e., $\angle PBA={100}^{\circ }$