O-LevelMathematicsTrigonometry4024/22 · May/June 2018 · Q79 Marks
A yacht sails the triangular route shown. The bearing of B from A is 135°. BC = 3.7km, AC = 2.8km and ABC = 42°. [Figure: triangle ABC with North line at A. Angle at A for bearing is 135 degrees from North. Side AC = 2.8km, BC = 3.7km, angle ABC = 42 degrees.]
- (a)Show that CÂB = 62.2°, correct to 1 decimal place.[3]
- (b)Find the bearing of A from C.[2]
- (c)Calculate the total length of the route.[4]
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Mark scheme
Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.
- [3] M2 M1 for 3.7 / sin CAB = 2.8 / sin 42 oe
- A1
- [2] M1 for 135 + 62.2 - 180 oe
- [4] B3 for 4.05 to 4.06
- OR M2 for √2.8^2 +3.7^2 - 2x2.8x3.7 × cos(180 – 42 – 62.2) oe or M1 for 2.8^2 +3.7^2-2×2.8×3.7×cos(180-42-62.2) oe
- OR M2 for 2.8 sin(180-42 – 62.2) / sin 42 oe or M1 for sin(180-42-62.2) / AB = sin 42 / 2.8 oe
- OR M2 for 3.7 sin(180-42 – 62.2) / sin 62.2 oe or M1 for sin(180-42-62.2) / AB = sin 62.2 / 3.7 oe
- OR B1 for ACB = 75.8
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