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

(a) 50° (b) 30° (c) 60° (d) 100°

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Best answer

(d) 100°

Explanation:

From ∆APB and ∆CPD,

∠APB = ∠CPD = 50° (since they are vertically opposite angles)

AP/PD = 6/5 … (i)

Also, BP/CP = 3/2.5

Or BP/CP = 6/5 … (ii)

From equations (i) and (ii),

We get,

AP/PD = BP/CP

So, ∆APB ∼ ∆DPC [using SAS similarity criterion]

∴ ∠A = ∠D = 30° [since, corresponding angles of similar triangles]

Since, Sum of angles of a triangle = 180°,

In ∆APB,

∠A + ∠B + ∠APB = 180°

So, 30° + ∠B + 50° = 180°

Then, ∠B = 180° – (50° + 30°)

∠B = 180 – 80° = 100°

Therefore, ∠PBA = 100°

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