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WAEC Mathematics 2011 Paper

Question : 37

Total: 47

Solution:

< Q =

\;\frac{240}{2}

(angle at centre twice that at the circumference)

< Q = 120^o

Also < POR = 360 - 240

= 120^o

( < s at centre) since /PQ/ = /QR/, < x = < R

Byt < x + < R + O + Q = 360 (sum of interior < s of quadrilateral)

x + R + 120 = 360^o

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}

= 60^o

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