Question

In the given 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 = ?

विकल्प

• 50°

• 30°

• 60°

• 100°

Answer 1

100°

Explanation:

PA·PC = 6 × 2.5 = 15 and PB·PD = 3 × 5 = 15, so PA·PC = PB·PD, therefore A, B, C, D lie on the same circle P is intersection of two chords.

∠CDP is an inscribed angle subtending arc CB, so arc CB = 2 × 30° = 60°.

The angle formed by the chords AP and BP at interior point P equals half the sum of the intercepted arcs:

`∠APB = 1/2(arc  AB + arc  CD) = 50^circ`, so arc AB + arc CD = 100°.

Hence, arc AD = 360° – (arc AB + arc BC + arc CD)

= 360° – (100° + 60°)

= 200°

The angle at B subtending arc AD (i.e. ∠PBA) is half that arc, so ∠PBA = `(200^circ)/2` = 100°.