Question

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

• A. ${50}^o$
• B. ${30}^o$
• C. ${60}^o$
• D. ${100}^o$

Answer 1

Answer: (D)

${100}^o$

Given: $PA=6$ cm, $PB=3$ cm, $PC=2.5$ cm, $PD=5$ cm
$\frac{PA}{PD}=\frac{6}{5}$

$\frac{PB}{PC}$ = $\frac{3}{2.5}$ = $\frac{6}{5}$

Thus, $\frac{PA}{PD}=\frac{PB}{PC}$

In $△APB$ and $△DPC$,
$\frac{PA}{PD}=\frac{PB}{PC}$ ...Given

$\angle APB=\angle DPC$ ...Vertically opposite angles

Hence, $△APB\sim △DPC$ ....S.A.S test of similarity

Hence, $\angle A=\angle D={30}^o$

Now, In $△APB$,
Sum of angles $={180}^o$
$\angle A+\angle B+\angle APB=180$
$30+50+\angle B={180}^o$
$\angle B={100}^{\circ }$