O-LevelMathematicsTrigonometry4024/22 · Oct/Nov 2016 · Q912 Marks
ABCD is a level playing field. AB = 65 m, BC = 70m and CA = 110m. CDA = 70°, DÂC = 58° and C is due South of B. [Figure showing a quadrilateral ABCD with given lengths and angles, and a North arrow]
- (a)Calculate the bearing of A from C.[4]
- (b)Calculate AD.[3]
- (c)(i)There are two vertical trees, AX and CY, each of height 17m, one at each end of the path AC. Calculate the angle of elevation of Y from B.[2]
- (c)(ii)A bird flies in a straight line from X to Y. It takes 24 seconds. Calculate the average speed of the bird. Give your answer in kilometres per hour.[3]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2016). The mark or mark type is shown in brackets.
- [4] 326 ft
- [2] M2 for 65² = 110² + 70² – 2×110×70× cos ACB soi or
- [1] M1 for the cosine rule with one error. and
- [1] A1 for 33.9 or 146.1 or 59.2 and
- [1] B1 ft for 360 – their ACB oe
- [2] SC 2 for 109.1 or 37.0
- [3] 92.2
- [2] M2 for AD/sin(70+58) or (180-(70+58)) = 110/sin 70 oe soi or
- [1] M1 for 70 + 58 or 180 – (70 + 58)
- [2] 13.6 or 13.7
- [1] M1 for tan YBC = 17/70 or tan BYC = 70/17
- [3] 16.5
- [1] M1 for Figs 110/24 soi and
- [1] B1 for × by 60×60/1000 oe soi
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