WAEC Mathematics Past Questions & Answers - Page 57

From the diagram,

h\(^2\) = 4\(^2\) + 3\(^2\) (pythagoras')

h\(^2\) = 16 + 9 = 25

h = \(\sqrt{25}\) = 5

Hence, sin x - cos x

= \(\frac{4}{5} - \frac{3}{5}\)

= \(\frac{1}{5}\)

Length of arc, L = 21.4 - 2 x 4.2cm

= 21.4 - 8.4

= 13cm

But L = \(\frac{\theta}{360^o}\) x 2\(\pi r\)

i.e 13 = \(\frac{\theta}{360^o}\)  x 2 x \(\frac{22}{7}\) x 4.2

= 13 x 360\(^o\) x 7

= \(\theta\) x 2 x 22 x 4.2

\(\theta\) = \(\frac{13 \times 360^o \times 7}{44 \times 4.2}\)

= \(\approx\) 177.27\(^o\)

\(\approx\) 177\(^o\) (to the nearest degree)

Total donation = 4 x 500 + 7 x 2000 + 20 x 1000 + 9 x 700 + 4 x 500 + 5 x 100 + 3 x 50 + 1 x 2 + 2 x 10 = 20000 + 14000 + 20000 + 6300 + 2000 + 500 + 150 + 2 + 20 = N62,972

m + n = 110\(^o\), (n + r) = 130\(^o\)

(m + n) = 120\(^o\)

then, r = 130\(^o\) - n

and;

m + (130^o - n) = 120\(^o\)

m - n = -10\(^o\)

2m + (n + r) = 110 + 120 = 230

2m + 130 = 230

2m = 230 - 130

m = \(\frac{100}{2}\) = 50\(^o\)

n = 110\(^o\) - 50\(^o\)

= 60\(^o\)

r = 130\(^o\) - 60\(^o\) = 70\(^o\)

Hence, the ratio m : n : r

= 50 : 60 : 70

= 5 : 6 : 7

In the diagram,

a = z (alternate angles)

b = 180\(^o\) - a (angles on a straight line)

b = 180\(^o\) - z

c = 180\(^o\) - x (angles on a straight line)

y = b + c (sum of oposite interior angles)

y = 180\(^o\) - z + 180\(^o\) - x

y = 360\(^o\) - z - x

x = 360\(^o\) - y - z