4MA0 Higher June 2014 Paper 4R Q19
19
Diagram NOT accurately drawn
\(A\), \(B\) and \(C\) are 3 villages.
\(B\) is 6.4 km due east of \(A\).
\(C\) is 3.8 km from \(A\) on a bearing of 210°
Calculate the bearing of \(B\) from \(C\).
Give your answer correct to the nearest degree.
Show your working clearly.
(6)
| Scheme | Marks |
|---|---|
| \((BC^2 =)\; 3.8^2 + 6.4^2 - 2 \times 3.8 \times 6.4 \cos 120^\circ\) (= 79.72) | M1 |
| \((BC^2 =)\) 14.44 + 40.96 + 24.32 (=79.72) | M1 A1 |
\(\dfrac{\sin C}{6.4} = \dfrac{\sin 120}{\text{“}8.92..\text{”}}\) or \(6.4^2 = 3.8^2 + \text{“}8.92\text{”}^2 - 2 \times 3.8 \times \text{“}8.92\text{”} \times \cos C\) | M1 |
\(\sin C = \dfrac{6.4 \times 0.866...}{\text{“}8.92..\text{”}}\) (= 0.62…) or \(\cos C = \dfrac{3.8^2 + \text{“}8.92\text{”}^2 - 6.4^2}{2 \times 3.8 \times \text{“}8.92\text{”}}\) (=0.78…) C = 38 – 38.5 | M1 |
| 068 | A1 |
| (6) | |
| (6 marks) |
Notes
M1: correct use of Cosine rule to find BC
M1: correct use of Sine rule or Cosine rule to find angle C
M1: correct rearrangement
A1: (0)68 – (0)68.4
Award M2 A1 for BC = 8.9 – 8.93 or \(\sqrt{79.72}\) or \(\sqrt{\dfrac{1993}{25}}\) oe
Award M2 for C = 38 – 38.5
Award M2 for B = 21.5 – 22 and C = 180 – 120 – B
Alternative
| Scheme | Marks |
|---|---|
| CD is the perpendicular from C to BA produced. \(\angle CAD = 60^\circ\) or \(ACD = 30^\circ\) | M1 |
| AD = 3.8 cos60° or 3.8sin30 (= 1.9) BD = 6.4+1.9 (= 8.3) | M1 A1 |
CD = 3.8sin60 or 3.8cos30 (=3.29) \(\tan BCD = \dfrac{8.3}{3.8\sin 60}\) oe | M1 M1 |
| 068 | A1 |
Notes
M1: uses triangle CAD and \(\angle CAD = 60^\circ\) or \(ACD = 30^\circ\)
CD may not be drawn in but can be implied
M1 A1: M1 for correct method to find horizontal length
A1 for BD = 8.3
A1: (0)68 – (0)68.4