(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°