2012
Mathematics
67ca00ce0c643c71d77dee1f
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: (90 – x)°
Explanation
< PQR = 90° (angle in a semi-circle) < QRP = (90 - x)°
< PQR = 90° (angle in a semi-circle) < QRP = (90 - x)°THIS WEEK's
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