2015 - WAEC Mathematics Past Questions and Answers - page 3

In the diagram given,

tan(\alpha) = (\frac{1}{0.5}) = 2

(\alpha) = tan - 1(2) = 63.43^o

= 63^o

In (\bigtriangleup)KBC, sin 30 = (\frac{800}{IKCI})

IKCL = (\frac{800}{sin30^o})

= (\frac{800}{0.5})

= 1600m

In a school with students' population 810, the sectoral angle for a class of 7 students is

= (\frac{72}{810}) x 360^o

= 32^o

Third quartile Q3 = (\frac{3}{4})Nth

= (\frac{3}{4}) x 20th score

= 15th score

= 63

Hence the probability that only Kebba will hit the target

= P(K)xP(E')xP(O')

= (\frac{2}{3} \times \frac{1}{4} \times \frac{1}{5})

= (\frac{1}{30})

In the diagram, (\frac{16}{12} = \frac{VX}{6}) (similar (\Delta)s)

VX = (\frac{16 \times 6}{12})

= 8cm

Let Tom's present agr be x.

Then x = 25 - n

If Tom is 5 years younger than Bade, then Bade's present age is x + 5 = 25 - n + 5

= (30 - n)

\(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3}}\) = \(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\)

= \(\frac{\sqrt{2} \times \sqrt{3} + \sqrt{3} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\)

= \(\frac{\sqrt{6} + 3}{3}\)

= \(\frac{3 + \sqrt{6}}{3}\)

= Hence, (m + n) = 1 + \(\frac{1}{3}\)


= 1\(\frac{1}{3}\)

In the diagram, QOR + 2m(vertically opposite angles)

So, m + 90° + 2m = 180°

(angles on str. line)

3m = 180° - 90°

3m = 90°

m = \(\frac{90^o}{3}\)

= 30°

substituting 30° for m in

2m + 4n = 200° gives

2 x 30° + 4n = 200°

60° + 4n = 200°

4n = 200° - 60°

= 140°

n = \(\frac{140°}{4}\)

= 35°