Question

In the figure, two line 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, \angle APB = 50°$ and $ \angle CDP = 30°$. Then, $ \angle PBA$ is equal to:

• A. $50°$
• B. $30°$
• C. $60°$
• D. $100°$

Answer 1

The correct option is D

$100°$

Step 1: Show that $△APB$ and $△DPC$ are similar

In the $△APB$ and $△DPC$, we can observe,

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

And, $\frac{BP}{CP}=\frac{3}{2.5}$

$\Rightarrow$ $\frac{BP}{CP}=\frac{3}{2.5}\times \frac{2}{2}$

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

From equations $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$, we get,

$\frac{AP}{PD}=\frac{BP}{CP}$ $...\left(\mathrm{iii}\right)$

Also, $\angle APB=\angle CPD=50°$ $...\left(\mathrm{iv}\right)$ [vertically opposite angles]

From equations $\left(\mathrm{iii}\right)$ and $\left(\mathrm{iv}\right)$, we get,

$∆APB~∆DPC$ [by SAS similarity axiom]

Step 2: Calculate the required angle

Now, since, $∆APB~∆DPC$

$\Rightarrow$ $\angle A=\angle D=30°$ [corresponding angles of similar triangles]

Now, applying the angle sum property in $△APB$, we have,

The sum of angles of the $△APB=180°$

$\Rightarrow$ $\angle APB+\angle PBA+\angle BAP=180°$

$\Rightarrow$ $50°+\angle PBA+30°=180°$

$\Rightarrow$ $\angle PBA+80°=180°$

$\Rightarrow$ $\angle PBA=180°-80°$

$\Rightarrow$ $\angle PBA=100°$

Hence, $\angle PBA=100°$, so, option (D) is the correct option.