M4 June 2011 Q4
4. A hiker walking due east at a steady speed of 5 km h\(^{-1}\) notices that the wind appears to come from a direction with bearing 050. At the same time, another hiker moving on a bearing of 320, and also walking at 5 km h\(^{-1}\), notices that the wind appears to come from due north.
Find
(a) the direction from which the wind is blowing, (3)
(b) the wind speed. (4)
| Scheme | Marks |
|---|---|
 | |
| 2 vector triangles with a common side | M1 |
| …..correct and drawn on a single diagram | A1 |
| Wind is from bearing \(025^\circ\), (N \(25^\circ\) E) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\dfrac{5}{\sin 25^\circ} = \dfrac{W}{\sin 40^\circ}\) (ft on their \(25^\circ\)) | M1 A1ft |
| \(W = \dfrac{5 \times \sin 40^\circ}{\sin 25^\circ} = 7.6\) (km h\(^{-1}\)) | M1 A1 |
| (4) | |
| (7 marks) |