In the diagram above, O is the center of the circle QOR is a diameter and ∠PSR is 37°. Find ∠PRQ
A.
37o
B.
53o
C.
65o
D.
127o
E.
147o
Correct answer is B
Construction: Join PQ.
Then < RSP = 37° = < RQP (angles on the same segment)
But < RPQ = 90° (angle in a semi-circle)
\(\therefore\) < PRQ = 180° - (90° + 37°)
= 53°