Numeracy Higher November 2023 Unit 2 Q9
9. A helicopter flies from Swansea to Neath and on from Neath to Port Talbot.
Diagram not drawn to scale
The bearing of Neath from Swansea is 073°.
The bearing of Port Talbot from Neath is 160°.
(a)
By calculation, show that the bearing of Swansea from Neath is 253°. [1]
(b)
From Port Talbot, the helicopter flies directly back to Swansea.
Calculate the bearing of this flight from Port Talbot to Swansea. [7]
The bearing of the flight from Port Talbot to Swansea is …………… °
From the formula list: In any triangle \(ABC\): Sine rule \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\); Cosine rule \(a^2 = b^2 + c^2 - 2bc\cos A\)
| Scheme | Marks | Comments |
|---|---|---|
| 160 + 20 + 73 or 180 + 73 or 180 – 73 = 107 AND 360 – 107 | B1 | Allow 160 + 93 |
| Scheme | Marks | Comments |
|---|---|---|
| (Distance Swansea to Port Talbot =) \(\sqrt{11^2 + 7.5^2 - 2 \times 11 \times 7.5 \times \cos 93(^\circ)}\) (=\(\sqrt{185.885\ldots}\)) | M2 | FT ‘their 93(°)’ M1 for 11\(^2\) + 7.5\(^2\) – 2×11×7.5×cos93(°) |
| = 13.6(339…) (km) | A1 | CAO. Mark final answer |
| (\(N\hat{S}P\) =) \(\sin^{-1}\left(\dfrac{\sin 93(^\circ)}{13.6(339\ldots)} \times 7.5\right)\) OR (\(N\hat{S}P\) =) \(\cos^{-1}\left(\dfrac{11^2 + 13.6(339\ldots)^2 - 7.5^2}{2 \times 11 \times 13.6(339\ldots)}\right)\) OR (\(N\hat{P}S\) =) \(\sin^{-1}\left(\dfrac{\sin 93(^\circ)}{13.6(339\ldots)} \times 11\right)\) OR (\(N\hat{P}S\) =) \(\cos^{-1}\left(\dfrac{7.5^2 + 13.6(339\ldots)^2 - 11^2}{2 \times 7.5 \times 13.6(339\ldots)}\right)\) | M2 | FT ‘their derived 13.6(339…)’ and ‘their 93(°)’ M1 for \(\dfrac{\sin \text{angle}}{7.5} = \dfrac{\sin 93(^\circ)}{13.6(339\ldots)}\) or equivalent OR M1 for 7.5\(^2\) = 11\(^2\)+13.6(339…)\(^2\)−2×11×13.6(339…)×cosNSP or equivalent OR M1 for \(\dfrac{\sin \text{angle}}{11} = \dfrac{\sin 93(^\circ)}{13.6(339\ldots)}\) or equivalent OR M1 for 11\(^2\) = 7.5\(^2\)+13.6(339…)\(^2\)−2×7.5×13.6(339…)×cosNPS or equivalent |
| (\(N\hat{S}P\) =) 33(.321…(°)) OR (\(N\hat{P}S\) =) 53.6(781…) or 53.7 or 54(°) | A1 | e.g. FT use of 13.6 leads to (\(N\hat{S}P\) =) 33.4(160…(°)) with (\(N\hat{P}S\) =) 53.8(732…(°)) |
| (Bearing =) 286(°) | A1 | Accept 286(.32…) (°) FT from second M2 only FT 180 + 73 + ‘their \(N\hat{S}P\)’ OR FT 360 – 20 – ‘their \(N\hat{P}S\)’ |