Question on: WAEC Mathematics - 2011
In the diagram, PO and OR are radii, |PQ| = |QR| and reflex < PQR is 240o. Calculate the value x
A
60o
B
55o
C
50o
D
45o
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Correct Option: A
< Q = \(\frac{240}{2}\) (angle at centre twice that at the circumference)
< Q = 120o
Also < POR = 360 - 240
= 120o
( < s at centre) since /PQ/ = /QR/, < x = < R
Byt < x + < R + O + Q = 360 (sum of interior < s of quadrilateral)
x + R + 120 = 360o
x + R = 360 - 240 = 120; Since x = R
x + x = 120
2x = 120
Since x = R
x + x = 120
2x = 120
x = \(\frac{120}{2}\)
= 60o
< Q = 120o
Also < POR = 360 - 240
= 120o
( < s at centre) since /PQ/ = /QR/, < x = < R
Byt < x + < R + O + Q = 360 (sum of interior < s of quadrilateral)
x + R + 120 = 360o
x + R = 360 - 240 = 120; Since x = R
x + x = 120
2x = 120
Since x = R
x + x = 120
2x = 120
x = \(\frac{120}{2}\)
= 60o
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