Numeracy Higher November 2019 Unit 2 Q7
7. Llinos walks to the summit of Snowdon, passing the lake called Llyn Glaslyn.
Her height above sea level increases by 485 m from Llyn Glaslyn to the summit.
From the summit, she sees two small boats on Llyn Glaslyn.
Both boats are in the same direction from the summit.
The angles of depression of the two boats are 41º and 27º, as shown in the diagram.
Diagram not drawn to scale
Calculate the distance between the boats. [5]
| Scheme | Marks | Comments |
|---|---|---|
| \(\dfrac{485}{\tan41(^\circ)}\) or tan49(º) × 485 or \(\dfrac{485 \times \sin49(^\circ)}{\sin41(^\circ)}\) OR \(\dfrac{485}{\tan27(^\circ)}\) or tan63(º) × 485 or \(\dfrac{485 \times \sin63(^\circ)}{\sin27(^\circ)}\) | M2 | M1 for \(\tan41(^\circ) = \dfrac{485}{d}\) or \(\tan49(^\circ) = \dfrac{d}{485}\) or \(\dfrac{d}{\sin49(^\circ)} = \dfrac{485}{\sin41(^\circ)}\) OR M1 for \(\tan27(^\circ) = \dfrac{485}{d}\) or \(\tan63(^\circ) = \dfrac{d}{485}\) or \(\dfrac{d}{\sin63(^\circ)} = \dfrac{485}{\sin27(^\circ)}\) |
| (Horizontal distances from Snowdon =) 557.9(…) or 558 (m) 951.8(…) or 951.9 or 952 (m) | A1 A1 | |
| (Distance between boats =) 393.8 to 394.1 (m) | B1 | FT ‘their 557.9(…) and ‘their 951.8(…) provided M1 or M2 awarded Answer needs to be correct for their values |
Alternative
Alternative method for final 3 marks:
| Scheme | Marks | Comments |
|---|---|---|
| Distance between boats =) 485(tan63 – tan49) OR \(485\left(\dfrac{1}{tan27} - \dfrac{1}{tan41}\right)\) | M2 | FT provided M2 previously awarded |
| (Distance between boats =) 393.8 to 394.1 (m) | A1 | Answer needs to be correct for their values |
Alternative
Alternative full method:
| Scheme | Marks | Comments |
|---|---|---|
| Correct use of sin or cos to find the lengths of 1 or 2 hypotenuse e.g. \(\dfrac{485}{\sin41}\) AND/OR \(\dfrac{485}{\cos63}\) (=739(.26…)) (=1068(.3…)) | M2 | Only award these M marks if they clearly go on to attempt to calculate the distance between the boats using the sine rule (only 1 needed) or cosine rule or 2 applications of Pythagoras (both needed) M1 for a correct expression for 1 of the hypotenuse if they go on to use the cosine rule or 2 applications of Pythagoras |
| \(\dfrac{739(.26…)}{\sin27}\) × sin14 OR \(\dfrac{1068(.3…)}{\sin139}\) × sin14 OR √(739(.26..)2+1068(.3..)2 – 2×739(.26..)×1068(.3..)×cos14) OR \(\sqrt{1068(.3..)^2 - 485^2} - \sqrt{739(.26\ldots)^2 - 485^2}\) | M2 | FT from M1 previously awarded if possible M1 for the unrearranged versions of the sine or cosine rule with correct substitution |
| (Distance between boats =) 393.8 to 394.1 (m) | A1 | CAO |
Corrected by Westie's Workshop
- The board's mark scheme printed “√(739(.26..)²+1068(.3..)² – 2×739(.26..)×(1068(.3..)²×cos14)”. Changed to “… – 2×739(.26..)×1068(.3..)×cos14)” because the cosine rule uses 2ab cos C, with each side once and not squared; the printed version also had an extra bracket. The corrected expression gives 393.9 m, as in the answer line.
Original: see the board's mark scheme PDF (link above).