Numeracy Higher November 2021 Unit 2 Q7
7. From its base on the Isle of Anglesey, an aircraft has to collect emergency hospital equipment from Birmingham to deliver to Cardiff.
Details of the journey are shown in the diagram below.
Diagram not drawn to scale
On what bearing did the aircraft fly from Birmingham to Cardiff?
Circle your answer. [1]
069° 135° 225° 235° 291°
From Cardiff, the aircraft flies directly back to its base on Anglesey.
The flight from Cardiff to Anglesey takes 30 minutes.
Calculate the average speed of the aircraft from Cardiff to Anglesey.
Give your answer in km/h. [5]
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 |
|---|---|---|
| 225º | B1 |
(i)
| Scheme | Marks | Comments |
|---|---|---|
| (AC =) \(\sqrt{215^2 + 165^2 - 2 \times 215 \times 165 \times \cos 69(^\circ)}\) (= \(\sqrt{48023(.7\ldots)}\)) | M2 | M1 for AC\(^2\) = 215\(^2\) + 165\(^2\) − 2×215×165×cos69(º) or M1 for AC\(^2\) = 48023(.7…) |
| = 219(.1…) (km) | A1 | |
| (Average speed =) \(\dfrac{219(.1\ldots)}{0.5}\) or equivalent | M1 | FT ‘their 219(.1…)’ provided sine/cosine rules attempted |
| = 438(·286…) (km/h) | A1 | Ignore a subsequent attempt to convert units |
(ii)
| Scheme | Marks | Comments |
|---|---|---|
| (BAC =) \(\sin^{-1}\left(\dfrac{\sin 69(^\circ) \times 165}{219(.1\ldots)}\right)\) or \(\cos^{-1}\left(\dfrac{215^2 + 219(.1\ldots)^2 - 165^2}{2 \times 215 \times 219(.1\ldots)}\right)\) | M2 | FT ‘their 219(.1…)’ If their speed from (b)(i) is used for ‘their 219(.1…)’ then FT for a possible M2A0M0A0 only M1 for
|
| = 44.6 to 44.7 (º) or 45 (º) | A1 | |
| (Bearing =) 360 – (180 – 114 – 44.6(6…))’ or 360 – (21.3(…) to 21.4) | M1 | FT ‘their derived 44.6(6…)’ Or full alternative method starting with the calculation of ACB |
| = 339 (º) | A1 | Allow 338.6(6…) to 338.7 (º) |
Alternative
7(b)(ii) Alternative method:
| Scheme | Marks | Comments |
|---|---|---|
| (ACB =) \(\sin^{-1}\left(\dfrac{\sin 69(^\circ) \times 215}{219(.1\ldots)}\right)\) or \(\cos^{-1}\left(\dfrac{165^2 + 219(.1\ldots)^2 - 215^2}{2 \times 165 \times 219(.1\ldots)}\right)\) | M2 | FT ‘their 219(.1…)’. If their speed from (b)(i) is used for ‘their 219(.1…)’ then FT for a possible M2A0M0A0 only M1 for
|
| = 66.3 to 66.43 (º) or 66 (º) | A1 | |
| (Bearing =) 360 – (66.3(…) – (114 – 69)) or 360 – (66.3(…) – 45) | M1 | FT ‘their derived 66.3(…)’ Or full alternative method starting with the calculation of BAC |
| = 339 (º) | A1 | Allow 338.6(6…) to 338.7 (º) |