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)
Explanation
(a) By cosine rule, (/PQ/^{2} = 20^{2} + 15^{2} – 2 times 20 times 15 times cos 145°)
= (400 + 225 – 600 cos 145°)
= (625 + 600 times 0.8192)
= (491.52 + 625)
(/PQ/^{2} = 1116.52 implies /PQ/ = sqrt{1116.52} = 33.41)
(approxeq 33km) ( to the nearest whole number)
(b) By sine rule,
(frac{sin < QPR}{15} = frac{sin 145}{33.41})
(sin < QPR = frac{15 sin 145}{33.41})
(sin < QPR = frac{15 times 0.5736}{33.41} = 0.2575)
(< QPR = 14.92°)
(therefore) Bearing of Q from P = 25° + 14.92° = 39.92°.
(approxeq 40°) (to nearest whole number)