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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, ∠APB=50∘ and ∠CDP=30∘. Then, ∠PBA is equal to

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

Answers (1)

 In △APB and △CPDAPPD=65BPCP=32.5=3025=65∠APB=∠CPD=50∘ {vertically opposite angles} ∴△APB∼△DPC{ {By SAS similarity criterion} ∴∠A=∠D=30∘{ corresponding angles of similar triangles}  In △APB∠A+∠B+∠APB=180∘ {Sum of interior angles of a triangle is 180∘}30∘+∠B+50∘=180∘∠B=180∘−50∘−30∘∠B=100∘⇒∠PBA=100∘
Hence option D is correct.

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infoexpert23

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