To solve the problem, we use the properties of similar triangles.
1. Understanding the Concept:
Given that $ \triangle ABC \sim \triangle PQR $, their corresponding angles are equal:
$ \angle A = \angle P $, $ \angle B = \angle Q $, $ \angle C = \angle R $
2. Given:
$ \angle P = 60^\circ $, $ \angle Q = 75^\circ $
Since $ \angle A = \angle P $,
$ \angle A = 60^\circ $
Final Answer:
The value of $ \angle A $ is $ {60^\circ} $

In the following figure \(\angle\)MNP = 90\(^\circ\), seg NQ \(\perp\) seg MP, MQ = 9, QP = 4, find NQ. 
Solve the following sub-questions (any four): In \( \triangle ABC \), \( DE \parallel BC \). If \( DB = 5.4 \, \text{cm} \), \( AD = 1.8 \, \text{cm} \), \( EC = 7.2 \, \text{cm} \), then find \( AE \). 