Higher January 2021 Paper 1 Q16

16 The diagram shows the positions of three ships, \(A\), \(B\) and \(C\).

Diagram NOT accurately drawn

Ship \(B\) is due north of ship \(A\).

The bearing of ship \(C\) from ship \(A\) is 120°

Calculate the bearing of ship \(C\) from ship \(B\).
Give your answer correct to the nearest degree.

(5)

Mark scheme

Scheme	Marks
\((BC^2 =)\;150^2 + 275^2 - (2 \times 150 \times 275 \times \cos 120)\) (= 139 375)	M1
\((BC =)\;\sqrt{150^2 + 275^2 + 41250}\) oe or \(\sqrt{139375}\) or \(25\sqrt{223}\) or 373.…	M1
e.g. \(\dfrac{\sin ABC}{275} = \dfrac{\sin 120}{\text{“}373...\text{”}}\) or \(275^2 = 150^2 + \text{“}373...\text{”}^2 - (2 \times 150 \times \text{“}373...\text{”} \times \cos ABC)\) or \(\cos ABC = \dfrac{150^2 + \text{“}373...\text{”}^2 - 275^2}{2 \times 150 \times \text{“}373...\text{”}}\) or \(\dfrac{\sin ACB}{150} = \dfrac{\sin 120}{\text{“}373...\text{”}}\) or \(150^2 = 275^2 + \text{“}373...\text{”}^2 - (2 \times 275 \times \text{“}373...\text{”} \times \cos ACB)\) or \(\cos ACB = \dfrac{275^2 + \text{“}373...\text{”}^2 - 150^2}{2 \times 275 \times \text{“}373...\text{”}}\)	M1
\((ABC =)\sin^{-1}\left(\dfrac{\sin 120}{\text{“}373...\text{”}} \times 275\right)\) (= 39.6…) or \((ABC =)\cos^{-1}\left(\dfrac{150^2 + \text{“}373...\text{”}^2 - 275^2}{2 \times 150 \times \text{“}373...\text{”}}\right)\) (= 39.6…) or \((ACB =)\sin^{-1}\left(\dfrac{\sin 120}{\text{“}373...\text{”}} \times 150\right)\) (= 20.3…) or \((ACB =)\cos^{-1}\left(\dfrac{275^2 + \text{“}373...\text{”}^2 - 150^2}{2 \times 275 \times \text{“}373...\text{”}}\right)\) (= 20.3…)	M1
Correct answer scores full marks (unless from obvious incorrect working) Answer: 140	A1
	(5)
	(5 marks)

M1: for correct substitution into the cosine rule

M1: for correct order of operations and square root

M1: (dep on 1^st M1) ft 373…

for a correct trig statement involving angle \(ABC\)
or angle \(ACB\)

M1: for a complete method to find angle \(ABC\) or angle \(ACB\)

A1: accept 140 – 140.4