n part (a), it was reported that although this question was attempted by majority of the
candidates, only those candidates who sketched the diagram correctly performed well on
the question.

The distance between the two ships could be obtained using either the sine or the cosine rule.

From the diagram, *∠*QHP = 900 - 420 = 480. This implies that *∠HQP *= *LQPH *= __180;48__ = 660. Using

2

the sine rule, __sin66°__ = __Sjin480__•Therefore, /PQ/ = __3~in4~O__ = 2.44km. Recall that LQPH = 66°.

*3 PQI s*in66°

Therefore, Bearing of Q from P = 270° + 66° = 336°.

In part (b), candidates' performance was reported to be poor. Candidates performed better in
part (a) than in part (b). They were expected to relate the time-taken during each part of the
journey with the total time for the whole journey. According to the report, their apparent
difficulty was in recalling the fact that average speed = __tota____l ____distance__ . Candidates wereexpecte to lnd t e time taken for t e whole Journey as

* total time taken *

__total____ distance covered__ = __600__ =10hrs.

*average speed * 60

lime taken when travelling with 75 krn/h = __300__ = 4hrs. Therefore, time taken when travelling

75

with v km/h = (10 - 4) hrs = 6hrs. i.e. __300__= 6. Hence, v = __300__ = 50km/h.

*v **6 *