This question was reported not to be popular with the candidates and the performance of those who attempted it was described as poor.
In part (a), candidates were expected to interpret the question correctly and apply the appropriate trigonometric ratio. However, poor application of the trigonometric ratios was observed by the Chief Examiner. Candidates were expected to show that ∠PQS = 60o which implied that ∠PQR = 30o. Hence, triangle PQR was isosceles and |QR| = 150 m. h would then be gotten using the relation  = cos 30o i.e. h = 150 x cos 30o = 130 m to the nearest whole number.

Candidates’ performance in part (b) was reported to be poorer than what it was in part (a). Candidates reportedly exhibited poor understanding of geometric theorems. Candidates were expected to show that ∠PQS = 360o – 252o) = 54o (angle at the centre of a circle is twice angle at the circumference). ∠QPS = 180o – 54o) = 63o (base angles of isosceles triangle PQS). ∠QSR = ∠QPS = 63o (angles in alternate segment). Hence, x + 63o + 79o = 180o (sum of angles in a triangle QSR), which gave x = 38o