General Mathematics Paper 2, May/June. 2011
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 Main
Weakness/Remedies
Strength

Question 8
1. Given that sin x = 0.6 and 0o ≤ × ≤ 90o, evaluate 2 cos x + 3 sin x, leaving your answer in the form  ,  where m and n are integers.

(b)

In the diagram, a semi-circle WXYZ with centre O is inscribed in an isosceles triangle ABC.              If |AC| =|BC|, |OC| = 30 cm and AĈB = 130o, calculate, correct to one decimal place, the

2. area of the shaded portion.

(Take π = )

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Observation

The Chief Examiner reported that although this question was not very popular, a good number of candidates who attempted it performed well in it.
In part (a), candidates were able to obtain the value of cos x as  . Hence 2cos x + 3sin x =      2( + 3() = .
In part (b), candidates were expected to recognize  and  as radii and that lines AC and BC were tangents to the semi-circle at points X and Y respectively. They were also expected to recall that tangents and radii were perpendicular at the point of contact. With this information, candidates were expected to show that |OX| = |OC| sin 65o = 30sin 65o = 27.7 cm. Area of semi-circle = (radius)2  =   (30sin 65o)2 = 1161.7cm2. Since the triangle is isosceles, then      ∠AOC = 90o and |OA| = |OB| = 30tan 65o. Therefore |AB| = 2  30tan 65o = 60 tan 65o. Area of the triangle = ½ base  height =   60tan65o 30 = 1930.1 cm2. Hence area of shaded part = 1930.1 – 1161.7 = 768.4 cm2