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)°
Обяснение
< PQR = 90° (angle in a semi-circle)
< QRP = (90 - x)°
In the diagram above, PQR is a circle centre O. If < QPR is x°, find < QRP.
< PQR = 90° (angle in a semi-circle)
< QRP = (90 - x)°