In the diagram above, PQR is a circle centre O. If < QPR is x^\circ, find < QRP.
Attached Figure / Diagram
A
x°
B
(90 – x)°
C
(90 + x)°
D
(180 – x)°
Verified Correct Answer: B.
Step-by-Step Mathematical Solution
< PQR = 90^\circ (angle in a semi-circle)
< QRP = (90 - x)^\circ
Source: Myschool Past Questions Archive (Record #33164)