2002 - WAEC Mathematics Past Questions and Answers - page 5

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 = \(10 + 10\sqrt{3}\)

= \(10(1 + \sqrt{3}) cm\)

Measured value = 0.375
Approximation = 0.4
Error = 0.4 - 0.375 = 0.025
Error% = (\frac{0.025}{0.375}) x 100% = 6.67% = 6.7%

(speed = \frac{distance}{time}<br /> 72 = \frac{60}{time}<br /> 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}<br /> speed \hspace{1mm}=60 \div \frac{2}{3} = 90km/hr)

Sum of the angles in a triangle = 180°

3x - 180° + 92° + x = 180°

4x - 88° = 180°

4x = 268°

x = 67°

< OSR = < ORS = \(\frac{180° - 64°}{2}\) = 58°

< PSR = 36° + 58° = 94°

< PSR + < PQR = 180°

94° + < PQR = 180° \(\implies\) < PQR = 180° - 94° = 86°

Given \(p = x - \frac{1}{x}\); \(q = x^2 + \frac{1}{x^2}\).

\(p^2 = (x - \frac{1}{x})(x - \frac{1}{x})\)

\(p^2 = x^2 + \frac{1}{x^2} - 2\)

\(p^2 = q - 2 \implies q = p^2 + 2\)

Number of at most 8 goals = 7

P(at most 8 goals) = \(\frac{7}{10}\)

< 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°