2013 - WAEC Mathematics Past Questions and Answers - page 5

Arts = 90^o business = 75^o, science = 45^o

Technical 360^o - (90^o - 75^o + 45^o) = 360^o - 210^o

Technical = 150^o

textbooks for technical class = \(\frac{150^o}{360^o}\) x 600

= 250

\(\frac{45^o}{360}\) x 100%

= 12\(\frac{1}{2}\)%

PQRST is a regular polygon with 5 sides-pentagon n = 5, Each exterior angle = \(\frac{360^o}{n}\)

= \(\frac{350}{5}\) = 70^o

Each interior angle = 180 - 72^o

(< s on str. line) = 180^o

Since the pentagon is regular, the exterior angles are equal < R = 72^o and < S = 72^o

< RVS = 180^o - (< R + < S)

= 180^o - (72^o + 72^o)

= 180^o - 144^o

= 36^o

Sum of angles on straight line = 180^o

80^o + 50^o + x + 10^o = 180^o

140^o + x = 180^o

x = 180^o - 140^o

x = 40^o

total sum of all angles = 360^o

2y^o + 2x^o + 80^o + 50^o + x + 10^o = 360^o

2y^o + 2(40)^o + 80^o + 50^o + 40 + 10 = 360^o

2y^o + 80 + 180^o = 360^o

2y^o + 260^o = 360^o

2y^o = 360^o - 260^o

2y^o = 100^o

angle 2y^o = 100^o

R +198 = 360(angles at a point)

R = 360 - 198 = 162^o

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 = 54^o

(x \(\cap\) y²) = [a, b]

n(x \(\cap\) y²) = 2