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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 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°            (D) 100°

Answers (1)

Answer: [D]

In \Delta APB and \Delta CPD.

\frac{AP}{PD}=\frac{6}{5}

\frac{BP}{CP}=\frac{3}{2.5}=\frac{30}{25}=\frac{6}{5}

\angle APB=\angle CPD=50^{o} {vertically opposite angles}

\therefore \Delta APB \sim \Delta DPC  {By SAS similarity criterion}

\therefore \angle A=\angle D=30^{o} {corresponding angles of similar triangles}

In \Delta APB

\angle A+\angle B+\angle APB=180^{o} {Sum of interior angles of a triangle is 180°}

30^{o}+\angle B+50^{o}=180^{o}

\angle B=180^{o}-50^{o}-30^{o}

\angle B=100^{o}

\Rightarrow \angle PBA=100^{o}

Hence option D is correct.

Posted by

infoexpert23

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