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