Numeracy Higher November 2022 Unit 2 Q8
8. Two cruise ships, The Explorer and The Magellan, leave the same port at 06:30.
The Explorer sails at a speed of 30 km/h on a bearing of 051°.
The Magellan sails at a speed of 35 km/h.
The angle between the courses of the two ships when they set sail is 49°.
At 11:00, The Magellan comes to a stop due to engine failure.
The diagram below shows the positions of the two ships at 11:00.
Diagram not drawn to scale
Immediately, The Explorer then heads directly towards The Magellan to pick up its passengers.
It travels at a speed of 30 km/h.
Calculate the time, to the nearest minute, when The Explorer will reach The Magellan. [7]
Time when The Explorer will reach The Magellan is ……………
Calculate the bearing The Explorer has to sail on from 11:00 onwards to arrive at The Magellan. [5]
From the formula list: In any triangle \(ABC\): Cosine rule \(a^2 = b^2 + c^2 - 2bc\cos A\)
| Scheme | Marks | Comments |
|---|---|---|
| (Distances travelled up to 11:00 =) 135 (km) AND 157.5 (km) | B1 | Check diagram |
| (Distance =) \(\sqrt{135^2 + 157.5^2 - 2 \times 135 \times 157.5 \times \cos 49(^\circ)}\) (=\(\sqrt{15132.33\ldots}\)) | M2 | FT their distances for M2 or M1 Allow use of 30 and 35 M1 for 135\(^2\) + 157.5\(^2\) − 2×135×157.5×cos49(º) |
| = 123(.01…) (km) | A1 | Must come from M2 and provided 30 and 35 not used in the cosine rule |
| (Time taken for Explorer to reach Magellan =) 123(.01…) ÷ 30 | M1 | Can only be awarded provided at least M1 previously awarded FT ‘their derived 123(.01…)’ |
| = 4.1(…) (hours) or 4 hrs 6 mins | A1 | |
| = 15:06 or 3:06 p.m. | A1 | FT from M1A0 for ‘their 4.1(…) (hours)’ provided of equivalent difficulty (not quarter or half hours involved) On FT, needs to be correct to the nearest minute, rounded or truncated If final M0A0A0 awarded, SC2 for an answer of 14:31 or 2:31 p.m. from the division by 35 OR SC1 for 3.5(1…) hours from the division by 35 |
| Scheme | Marks | Comments |
|---|---|---|
| (Angle at top of triangle =) \(\sin^{-1}\left(\dfrac{\sin 49(^\circ)}{123(.01\ldots)} \times 157.5\right)\) OR \(\cos^{-1}\left(\dfrac{135^2 + 123(.01\ldots)^2 - 157.5^2}{2 \times 135 \times 123(.01\ldots)}\right)\) | M2 | FT their values consistently used from (a) M1 for \(\dfrac{\sin \text{angle}}{157.5} = \dfrac{\sin 49(^\circ)}{123(.01\ldots)}\) or equivalent OR M1 for 157.5\(^2\) = 135\(^2\)+123(.01)\(^2\) – 2×135×123(.01)×cos angle |
| 75(.08…) to 75.105 (º) | A1 | Must come from M2 |
| (Bearing =) 360 – (180 – 51) – 75(.08…) or 180 – (75(.08…) – 51) or 231 – 75(.08…) | M1 | FT ‘their derived 75(.08…)’ |
| = 156 (º) | A1 | Allow an answer of 155.9(19…)(º) |
Alternative
8(b) Alternative method:
| Scheme | Marks | Comments |
|---|---|---|
| (Angle at right of triangle =) \(\sin^{-1}\left(\dfrac{\sin 49(^\circ)}{123(.01\ldots)} \times 135\right)\) OR \(\cos^{-1}\left(\dfrac{157.5^2 + 123(.01\ldots)^2 - 135^2}{2 \times 157.5 \times 123(.01\ldots)}\right)\) | M2 | FT their values consistently used from (a) M1 for \(\dfrac{\sin \text{angle}}{135} = \dfrac{\sin 49(^\circ)}{123(.01\ldots)}\) or equivalent OR M1 for 135\(^2\)=157.5\(^2\)+123(.01)\(^2\) –2×157.5×123(.01)×cos angle |
| = 55.9(19…) to 56 (º) | A1 | Must come from M2 |
| (Bearing =) 360 – (180 – 51) – (180 – 49 – 55.9(19…)) or 51 + 49 + 55.9(19…) or 100 + 55.9(19…) | M1 | FT ‘their derived 55.9(19…)’ |
| = 156 (º) | A1 | Allow an answer of 155.9(19…)(º) |