Higher June 2019 Paper 1R Q25
25 A boat sails from point \(X\) to point \(Y\) and then to point \(Z\).
\(Y\) is on a bearing of 280° from \(X\).
\(Z\) is on a bearing of 220° from \(Y\).
The distance from \(X\) to \(Y\) is 3.5 km.
The distance from \(Y\) to \(Z\) is 6 km.
Work out the bearing of \(Z\) from \(X\).
Give your answer correct to 1 decimal place.
(5)
| Scheme | Marks |
|---|---|
| e.g. (220 – 180) + (360 – 280) (= 120) | M1 |
| \(XZ = \sqrt{3.5^2 + 6^2 - 2 \times 3.5 \times 6 \times \cos(\text{``}{120}\text{''})}\) (=8.3… or \(\dfrac{\sqrt{277}}{2}\)) | M1 |
| \(\dfrac{\sin YXZ}{6} = \dfrac{\sin(\text{``}{120}\text{''})}{\text{``}{8.32…}\text{''}}\) | M1 |
| \(YXZ = \sin^{-1}\left(\dfrac{6\sin(\text{``}{120}\text{''})}{\text{``}{8.32…}\text{''}}\right)\) (=38.6…) | M1 |
| 241.4 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for a method to find angle \(XYZ\).
Could be seen on a diagram
M1: or \(6^2 = 3.5^2 + \text{``}{8.32}\text{''}^2 - 2 \times 3.5 \times \text{``}{8.32}\text{''} \times \cos YXZ\)
M1: or \(YXZ = \cos^{-1}\left(\dfrac{3.5^2 + \text{``}{8.32}\text{''}^2 - 6^2}{2 \times 3.5 \times \text{``}{8.32}\text{''}}\right)\) (= 38.6…)
A1: accept 241.2 – 241.4