In the adjoining 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

1. 50°

2. 30°

3. 60°

4. 100°

Considering △APB and △CPD,

$\dfrac{AP}{PD} = \dfrac{6}{5} \text{ and } \dfrac{BP}{CP} = \dfrac{3}{2.5} = \dfrac{6}{5}.$ and ∠APB = ∠CPD (Vertically opposite angles are equal)

∴ △APB ~ △CPD

Hence, ∠PAB = ∠PDC = 30°

∠PBA = 180° - (∠PAB + ∠APB) = 180° - (30° + 50°) = 180° - 80° = 100°.

Hence, Option 4 is the correct option.