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  • In Fig., 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^{o} and \angle CDP=30^{o}. Then, \angle PBA is equal to (A) 50° (B) 30°(C) 60°

In Fig., two line segments $A C$ and $B D$ intersect each other at the point $P$ such that $P A=6 \mathrm{~cm}, P B=3 \mathrm{~cm}, P C=2.5 \mathrm{~cm}, P D=5 \mathrm{~cm}$, $\angle A P B=50^{\circ}$ and $\angle C D P=30^{\circ}$. Then, $\angle P B A$ is equal to

(A) 50°            (B) 30°            (C) 60°            (D) 100°

Answers (1)

$
\begin{aligned}
& \text { In } \triangle A P B \text { and } \triangle C P D \\
& \frac{A P}{P D}=\frac{6}{5} \\
& \frac{B P}{C P}=\frac{3}{2.5}=\frac{30}{25}=\frac{6}{5} \\
& \angle A P B=\angle C P D=50^{\circ} \text { \{vertically opposite angles\} } \\
& \therefore \triangle A P B \sim \triangle D P C\{\text { \{By SAS similarity criterion\} } \\
& \therefore \angle A=\angle D=30^{\circ}\{\text { corresponding angles of similar triangles\} } \\
& \text { In } \triangle A P B \\
& \left.\angle A+\angle B+\angle A P B=180^{\circ} \text { \{Sum of interior angles of a triangle is } 180^{\circ}\right\} \\
& 30^{\circ}+\angle B+50^{\circ}=180^{\circ} \\
& \angle B=180^{\circ}-50^{\circ}-30^{\circ} \\
& \angle B=100^{\circ} \\
& \Rightarrow \angle P B A=100^{\circ}
\end{aligned}
$
Hence option D is correct.

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infoexpert23

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