2002 - WAEC Mathematics Past Questions and Answers - page 5
In the diagram, |QR| = 10cm, PR⊥QS, angle PSR = 30° and angle PQR = 45°. Calculate in meters |QS|
In \(\Delta\) PQR,
\(\tan 45 = \frac{PR}{10} \implies PR = 10 \tan 45\)
= 10m
In \(\Delta\) PRS,
\(\tan 30 = \frac{10}{RS} \implies RS = \frac{10}{\tan 30}\)
= \(\frac{10}{\frac{1}{\sqrt{3}}\)
= \(10\sqrt{3}\)
PS = \(10 + 10\sqrt{3}\)
= \(10(1 + \sqrt{3}) cm\)
Approximation = 0.4
Error = 0.4 - 0.375 = 0.025
Error% = \(\frac{0.025}{0.375}\) x 100% = 6.67% = 6.7%
72 = \frac{60}{time}\
t = \frac{60}{72} = \frac{5}{6}hr\)
time lost = 10mis \(= \frac{10}{60}hr = \frac{1}{6}\)
Time required for the journey
\(=\frac{5}{6}-\frac{1}{6} = \frac{2}{3}\
speed \hspace{1mm}=60 \div \frac{2}{3} = 90km/hr\)
In the diagram, \(QR||TP and W\hat{P}T = 88^{\circ} \). Find the value of x
Sum of the angles in a triangle = 180°
3x - 180° + 92° + x = 180°
4x - 88° = 180°
4x = 268°
x = 67°
In the diagram O is the center of the circle, ∠SOR = 64° and ∠PSO = 36°. Calculate ∠PQR
< OSR = < ORS = \(\frac{180° - 64°}{2}\) = 58°
< PSR = 36° + 58° = 94°
< PSR + < PQR = 180°
94° + < PQR = 180° \(\implies\) < PQR = 180° - 94° = 86°
Given that \(p = x-\frac{1}{x} and\hspace{1mm}q = x^2 + \frac{1}{x^2}\) express q in terms of p.
Given \(p = x - \frac{1}{x}\); \(q = x^2 + \frac{1}{x^2}\).
\(p^2 = (x - \frac{1}{x})(x - \frac{1}{x})\)
\(p^2 = x^2 + \frac{1}{x^2} - 2\)
\(p^2 = q - 2 \implies q = p^2 + 2\)
The number of goals scored by a school team in 10 netball matches are as follows: 3, 5, 7, 7, 8, 8, 8, 11, 11, 12. Find the probability that in a match, the school team will score at most 8 goals.
Number of at most 8 goals = 7
P(at most 8 goals) = \(\frac{7}{10}\)
In the diagram, LMT is a straight line. lf O is the centre of circle LMN, OMN = 20°, LTN = 32° and |NM| = |MT|, find LNM.
< MNT = < MTN = 32°
< NMT = 180° - 2(32°) = 116°
< OMN + < NMT + < LMO = 180°
20° + 116° + < LMO = 180° \(\implies\) < LMO = 180° - 136° = 44°
< LMN = < LMO + < OMN
= 44° + 20° = 64°
< NOM = 180° - 2(20°) = 140°
< NLM = \(\frac{1}{2} \times < NOM = 70°\)
< LNM + < LMN + < NLM = 180°
< LNM + 64° + 70° = 180°
< LNM = 180° - 134° = 46°