(d) 100°
Explanation:
From ∆APB and ∆CPD,
∠APB = ∠CPD = 50° (since they are 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),
We get,
AP/PD = BP/CP
So, ∆APB ∼ ∆DPC [using SAS similarity criterion]
∴ ∠A = ∠D = 30° [since, corresponding angles of similar triangles]
Since, Sum of angles of a triangle = 180°,
In ∆APB,
∠A + ∠B + ∠APB = 180°
So, 30° + ∠B + 50° = 180°
Then, ∠B = 180° – (50° + 30°)
∠B = 180 – 80° = 100°
Therefore, ∠PBA = 100°