In the diagram above, PQR is a circle centre O. If < QPR is x°, find < QRP.
A.
x°
B.
(90 – x)°
C.
(90 + x)°
D.
(180 – x)°
Correct answer is B
< PQR = 90° (angle in a semi-circle) < QRP = (90 - x)°