In the diagram above, PR is perpendicular from P to QS, PQ = 2cm, QR = 1cm and PR = RS. what is the size of the angle QPS?
A.
135o
B.
105o
C.
90o
D.
75o
E.
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°
\(\therefore < QPS = 30° + 45° = 75°\)