135o
105o
90o
75o
60o
Correct answer is D
\(\Delta RPS\) = 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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