(a) 50° (b) 30° (c) 60° (d) 100°

Solution:

(d) 100°

Clarification:

From ∆APB and ∆CPD,

\[\angle APB\text{ }=\angle CPD\text{ }=\text{ }50{}^\circ \] (since they are upward inverse points)

\[AP/PD\text{ }=\text{ }6/5\text{ }\ldots \text{ }\left( I \right)\]

Likewise, BP/CP = 3/2.5

Or then again BP/CP = 6/5 … (ii)

From conditions (I) and (ii),

We get,

\[AP/PD\text{ }=\text{ }BP/CP\]

In this way, ∆APB ∼ ∆DPC [using SAS closeness criterion]

\[\therefore \angle A\text{ }=\angle D\text{ }=\text{ }30{}^\circ \]  [since, relating points of comparative triangles]

Since, Sum of points of a triangle = 180°,

In ∆APB,

\[\begin{array}{*{35}{l}}

   \angle A\text{ }+\angle B\text{ }+\angle APB\text{ }=\text{ }180{}^\circ   \\

   ~  \\

\end{array}\]

In this way, \[30{}^\circ \text{ }+\angle B\text{ }+\text{ }50{}^\circ =\text{ }180{}^\circ \]

Then, at that point, ∠B = 180° – (50° + 30°)

\[\angle B\text{ }=\text{ }18080{}^\circ =\text{ }100\]

Hence, ∠PBA = 100°