General Mathematics Paper 2,Nov/Dec. 2010  
Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 Main
General Comments

  1. From a point X, a boat sails 6 km on a bearing of 037o to a point Y. It then sails 7 km from Y on a bearing of 068o to a point Z. Calculate the:

    1. distance XZ, correct to two decimal places;
    2. bearing of Z from X, correct to the nearest degree.


It was also reported that this question was well attempted by majority of the candidates. However, many of them did not draw the correct diagram. Some did not draw any diagram at all. It was also observed that while some candidates were able to calculate ∠YXZ correctly to obtain 16.72o, they did not add this value to 37o to obtain the required bearing. Others did not write the bearing as whole numbers. Candidates were expected to represent the information in a diagram like the one below.






7 km




6 km









Using the diagram, XŶZ = 37o +90o + 22o. Applying the cosine rule, |XZ|2 = 72 + 62 – (7)(6)cos 149o = 157.002. |XZ| =  = 12.53 km. If ∠YXZ = x, then using the sine rule,   . Simplifying this equation gave x = 16.72o. The bearing of Z from X would then be 037o+ 016.72o = 054o or N54oE.

Questions 14 and 15 were not meant for candidates in Nigeria and responses were not received from them.


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