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°

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