(a) Given that 5cos ( + 8.5)o – 1 = 0, 0o££ 90o, calculate, correct to the nearest degree, the value of .
(b) The bearing of Q from P is 150o and the bearing of P from R is 015o. If Q and R are
24 km and 32 km respectively from P:
represent this information in a diagram;
calculate the distance between Q and R, correct to two decimal places;
find the bearing of R from Q, correct to the nearest degree.
The Chief Examiner reported that this question was quite popular among the candidates. Their performance was described as fair.
In part (a), it was observed that when solving the trigonometric equation 5cos(x + 8.5)o = 1, some candidates expanded the bracket to obtain This was a wrong method. Candidates were expected to respond as follows: Dividing both sides of the sign of equality by 5 gave0.2. This implied that=Therefore. =(78.46-8.5)o=70correct to the nearest degree.
In part (b), majority of the candidates were reported not to draw the diagram. Some others did not draw the correct diagram. Candidates were encouraged to draw diagrams in those question where diagrams would aid them in answering the question. Teachers were encouraged to stress this during instruction.
Candidates were expected to draw the following diagram
From the diagram, ∠. Using the cosine rule, the distance QR would be obtained as follows: │QR│222. Therefore │QR│ = = 22.67 km. Angle PQR was found using the sine rule as∠ Therefore, bearing of R from Q was.
+ 8.5)o – 1 = 0, 0o£
£ 90o, calculate, correct to the nearest degree, the value of 
.
This was a wrong method. Candidates were expected to respond as follows:
Dividing both sides of the sign of equality by 5 gave
0.2. This implied that
=
Therefore. 
=(78.46-8.5)o=70
correct to the nearest degree.
. Using the cosine rule, the distance QR would be obtained as follows: │QR│2
2
2
. Therefore │QR│ = 
= 22.67 km. Angle PQR was found using the sine rule as∠
Therefore, bearing of R from Q was
.