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JAMB - Mathematics (2012 - No. 42)

In the diagram above, PQR is a circle centre O. If < QPR is x°, find < QRP.

x°
(90 – x)°
(90 + x)°
(180 – x)°

Penjelasan

< PQR = 90° (angle in a semi-circle)

< QRP = (90 - x)°

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