The Chief Examiner reported that majority of the candidates who attempted this question did not illustrate the given information in a diagram correctly. This also affected their being able to solve the question correctly. Candidates were expected to represent the information in a diagram as shown:

Q 140o

8 km
20km R
r
o
P

From the diagram, ∠PQR = 180o – 140o = 40o. Using the cosine rule,
|PR|2 = 202 + 82 – 2(20)(8) cos 40o simplifying this equation gave
|PR| = 14.79411km. Using the sine rule = where is the bearing of R from P. Solving this equation gave = 20.34o i.e. bearing of R from P is 020o or N20oE. Let r be the distance of R north of P. Then r = |PR| cos = 13 km, correct to 2 significant figures.