2013 - WAEC Mathematics Past Questions and Answers - page 5
Arts = 90o business = 75o, science = 45o
Technical 360o - (90o - 75o + 45o) = 360o - 210o
Technical = 150o
textbooks for technical class = (\frac{150^o}{360^o}) x 600
= 250
PQRST is a regular polygon with 5 sides-pentagon n = 5, Each exterior angle = (\frac{360^o}{n})
= (\frac{350}{5}) = 70o
Each interior angle = 180 - 72o
(< s on str. line) = 180o
Since the pentagon is regular, the exterior angles are equal < R = 72o and < S = 72o
< RVS = 180o - (< R + < S)
= 180o - (72o + 72o)
= 180o - 144o
= 36o
Sum of angles on straight line = 180o
80o + 50o + x + 10o = 180o
140o + x = 180o
x = 180o - 140o
x = 40o
total sum of all angles = 360o
2yo + 2xo + 80o + 50o + x + 10o = 360o
2yo + 2(40)o + 80o + 50o + 40 + 10 = 360o
2yo + 80 + 180o = 360o
2yo + 260o = 360o
2yo = 360o - 260o
2yo = 100o
angle 2yo = 100o
R +198 = 360(angles at a point)
R = 360 - 198 = 162o
C = 72(alternate angle), b + c = 180(angles or straight line)
b = 190 - c = 180 - 72 = 108
a + b = R = 162
a = R - b = 162 - 108
= 54, but a = y(alternate angles)
y = 54o