General Mathematics Paper 2, May/June. 2010
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 Main
Weakness/Remedies
Strength

Question 9

In the diagram, /AB/ = 8 km, /BC/ = 13 km, the bearing of A from B is 310° and the bearing of B from C is 230°. Calculate, correct to 3 significant figures,
(a) the distance AC;
(b) the bearing of C from A;
(c) how far east of B, C is

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Observation

This question was reported to have been attempted by majority of the candidates and their performance was described as satisfactory. However, most of them were reported not to calculate LABC correctly hence got wrong answers even though they were able to apply the cosine rule correctly to their wrong values. Others were not able to determine the required bearing correctly. The expected responses were as follows:
LASC = 100°. Therefore /AC/ = 82 + 132 - 2 (8)(13)cos100 which gave / AC/ = 16.4 km.
sin (LCAS) = sin100. Hence, sin (LCAB) = 13Sin100
13                  16.4                                            16.4
Simplifying gave LCAB = 51.32°. Bearing of C from A = 180 - (50 + 51.32) = 079°.
If the distance of C east of B = BD, then BO = BC cas 40° = 13 x cas 40° = 9.96 km.