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
In ∆APB and ∆CPD,
∠APB = ∠CPD = 50° (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)
AP/PD = BP/CP
∴ ∆APB ∼ ∆DPC [by SAS similarity criterion]
∴ ∠A = ∠D = 30° [corresponding angles of similar triangles]
In ∆APB,
∠A + ∠B + ∠APB = 180°
[Sum of angles of a triangle = 180°]
⇒ 30° + ∠B + 50° = 180°
∴ ∠B = 180° - (50° + 30°)
∠B = 180 – 80° = 100°
∠PBA = 100°