Two men P and Q set off from a base camp R,... | WAEC SSCE - General Mathematics - 1996 (Essay)

Question 1 Report

Two men P and Q set off from a base camp R, prospecting for oil. P moves 20km on a bearing of 205° and Q moves 15km on a bearing of 060°. Calculate the:

(a) distance of Q from P ;

(b) bearing of Q from P.

(Give your answer in each case to the nearest whole number)

Answer

Place base camp R at the origin. Convert each bearing to \((\text{East}, \text{North})\) components using East \(= r\sin\theta\), North \(= r\cos\theta\).

P: 20 km on bearing \(205^\circ\):

\[ P = (20\sin 205^\circ,\ 20\cos 205^\circ) = (-8.45,\ -18.13). \]

Q: 15 km on bearing \(060^\circ\):

\[ Q = (15\sin 60^\circ,\ 15\cos 60^\circ) = (12.99,\ 7.50). \]

(a) Distance of Q from P. The displacement is

\[ Q - P = (12.99 + 8.45,\ 7.50 + 18.13) = (21.44,\ 25.63). \] \[ PQ = \sqrt{21.44^2 + 25.63^2} = \sqrt{459.7 + 656.9} = \sqrt{1116.6} = 33.4 \approx 33\ \text{km}. \]

(Check by the cosine rule with included angle \(205^\circ - 60^\circ = 145^\circ\): \(PQ^2 = 20^2 + 15^2 - 2(20)(15)\cos 145^\circ = 1116.5\), giving \(PQ = 33\) km.)

(b) Bearing of Q from P. Both components of \(Q - P\) are positive, so Q is north-east of P:

\[ \tan\theta = \frac{\text{East}}{\text{North}} = \frac{21.44}{25.63} = 0.8366 \implies \theta = 39.9^\circ. \]

The bearing of Q from P is \(\approx 040^\circ\).

Read lesson note on Bearings (WAEC)

Read lesson note on Circles (WAEC)

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