2015 - WAEC Mathematics Past Questions and Answers - page 3
tan\(\alpha\) = \(\frac{1}{0.5}\) = 2
\(\alpha\) = tan - 1(2) = 63.43o
= 63o
IKCL = \(\frac{800}{sin30^o}\)
= \(\frac{800}{0.5}\)
= 1600m
= \(\frac{72}{810}\) x 360o
= 32o
= \(\frac{3}{4}\) x 20th score
= 15th score
= 63
Scores & 0 - 4 & 5 - 9 & 10 - 14\
\hline Frequency & 2 & 1 & 2\end{array}\)
The table shows the distribution of the scores of some students in a test. Calculate the mean scores.
= P(K)xP(E')xP(O')
= \(\frac{2}{3} \times \frac{1}{4} \times \frac{1}{5}\)
= \(\frac{1}{30}\)
VX = \(\frac{16 \times 6}{12}\)
= 8cm
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)
If \(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3}}\) is simplified as m + n\(\sqrt{6}\), find the value of (m + 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 given diagram, \(\bar{QT}\) and \(\bar{PR}\) are straight lines, < ROS = (3n - 20), < SOT = n, < POL = m and < QOL is a right angle. Find the value of n.
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°