Q26 of 180 Page 2

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


In ∆APB and ∆CPD,

∠APB = ∠CPD = 50° (vertically opposite angles)


…(i)


Also,


Or …(ii)


From equations (i) and(ii)



∴ ∆APB ∼ ∆DPC [by SAS similarity criterion]


∴∠A = ∠D = 30° [corresponding angles of similar triangles]


In ∆APB,


∠BAP + ∠PBA + ∠APB = 180° [Sum of angles of a triangle = 180°]


30° + ∠PBA + 50° = 180°


∴∠PBA = 180° – (50° + 30°)


∠PBA = 180 – 80° = 100°


∠PBA = 100°


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