Question

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 ______.

Options

• 50°

• 30°

• 60°

• 100°

Answer 1

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 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°