In the given figure, two line segments AC and BD intersect each other at point P such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°. Then ∠PBA = ?

1. 30°

2. 50°

3. 60°

4. 100°

Considering ΔAPB and ΔCPD,

$\dfrac{AP}{PD} = \dfrac{6}{5} \text{ and } \dfrac{BP}{CP} = \dfrac{3}{2.5} = \dfrac{6}{5}$

∠APB = ∠CPD [Vertically opposite angles are equal]

∴ ΔAPB ∼ ΔDPC (By SAS similarity)

Hence, ∠PAB = ∠PDC = 30°

⇒ ∠PBA = 180° - (∠PAB + ∠APB)

⇒ ∠PBA = 180° - (30° + 50°)

⇒ ∠PBA = 180° - 80°

⇒ ∠PBA = 100°.

Hence, option 4 is the correct option.