2018 - WAEC Mathematics Past Questions and Answers - page 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)

From the diagram,

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

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

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

Hence, sin x - cos x

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

= \(\frac{2}{5}\)

In \(\bigtriangleup\)YSC, sin 30\(^o\) = \(\frac{YS}{20}\)

|YS| = 20 sin 30\(^o\)

= 20 x 0.5

10m

Let the length of a side of the rhombus be n

Then, n\(^2\) = 5\(^2\) + 12\(^2\)

= 25 + 144 = 169

n = \(\sqrt{169}\)

= 13cm

Hence, perimeter of rhombus = 4n = 4 x 13

= 52cm

Let n(M \(\cup\) N \) = x

Then 20 - x + x + 30

- x = n(M \(\cup\) N)

50 - x = 40

50 - 40 = x

10 = x

x = 10

Hence, n(M \(\cup\N)' = 8  + (20 - 10) + (30 + 10)

= 8 + 10 + 20

= 38

y = x\(^2\) ....(1)

y = x ......(2)

y = y

x\(^2\) - x

x\(^2\) - x = 0

x(x - 1) = 0

x = 0 or x - 1 = 0

x = 0 or x = 1

when x = 0, y = 0\(^2\) = 0

when x = 1, y = 1\(^2\) = 1

Hence; the two graphs interest at (0, 0) and (1, 1)