Foundation June 2019 Paper 2 Q21
21 The diagram shows the positions of three ships \(A\), \(B\) and \(C\).
\(B\) is 234 km due north of \(A\).
\(C\) is 356 km due east of \(A\).
Work out the bearing of \(B\) from \(C\).
Give your answer correct to the nearest degree.
(4)
| Scheme | Marks |
|---|---|
\(\tan(BCA) = \dfrac{234}{356}\) or \(\tan(ABC) = \dfrac{356}{234}\) | M1 |
(\(BCA =\)) \(\tan^{-1}\left(\dfrac{234}{356}\right)\) (= 33(.317…)) or (\(ABC =\)) \(\tan^{-1}\left(\dfrac{356}{234}\right)\) (= 56(.682…)) | M1 |
| 270 + “33.317…” or 360 − “56.682…” | M1 |
| 303 | A1 |
| (4) | |
| (4 marks) |
Notes
M1: candidates that use Pythagoras to find \(BC\) can gain M1 for
\(\cos BCA = \dfrac{356}{\sqrt{234^2 + 356^2}}\) or \(\sin BCA = \dfrac{234}{\sqrt{234^2 + 356^2}}\) or
\(\sin ABC = \dfrac{356}{\sqrt{234^2 + 356^2}}\) or \(\cos ABC = \dfrac{234}{\sqrt{234^2 + 356^2}}\)
M1: (dep) complete method to find angle \(BCA\) or \(ABC\)
M1: complete method
A1: accept 303 – 303.4