135o
105o
90o
75o
60o
Correct answer is D
ΔRPS = Isosceles triangle
\therefore < RPS = \frac{180° - 90°}{2}
= 45°
In \Delta QPR,\sin < QPR = \frac{1}{2}
= 30°
\therefore < QPS = 30° + 45° = 75°
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