Higher November 2024 Paper 2 Q24
24 \(A\), \(B\) and \(C\) are three points on horizontal ground.
\(AB = 8.4\) metres \(BC = 9.2\) metres
\(B\) is on a bearing of 067° from \(A\)
\(C\) is on a bearing of 129° from \(B\)
Calculate the bearing of \(A\) from \(C\)
Give your answer correct to the nearest degree.
(6)
| Scheme | Marks |
|---|---|
| 67 + 51 (= 118) or angle split into 67 and 51 | M1 |
| \(AC^2 = 8.4^2 + 9.2^2 - 2 \times 8.4 \times 9.2 \times \cos\text{``}{118}\text{''}\;(= 227.(7615))\) (227 – 228) | M1 |
\(AC = \sqrt{8.4^2 + 9.2^2 - 2 \times 8.4 \times 9.2 \times \cos\text{``}{118}\text{''}}\;(= 15.(09\ldots))\) (15 – 15.1) | M1 |
\(\dfrac{\sin ACB}{8.4} = \dfrac{\sin\text{``}{118}\text{''}}{\text{``}{15.09\ldots}\text{''}}\) or \(\dfrac{\sin BAC}{9.2} = \dfrac{\sin\text{``}{118}\text{''}}{\text{``}{15.09\ldots}\text{''}}\) or \(\cos ACB = \dfrac{9.2^2 + \text{``}{15.09}\text{''}^2 - 8.4^2}{2 \times 9.2 \times \text{``}{15.09}\text{''}}\) or \(\cos BAC = \dfrac{8.4^2 + \text{``}{15.09}\text{''}^2 - 9.2^2}{2 \times 8.4 \times \text{``}{15.09}\text{''}}\) | M1 |
\((ACB =)\sin^{-1}\left(\dfrac{8.4\sin\text{``}{118}\text{''}}{\text{``}{15.09\ldots}\text{''}}\right)\) or \(\cos^{-1}\left(\dfrac{9.2^2 + \text{``}{15.09}\text{''}^2 - 8.4^2}{2 \times 9.2 \times \text{``}{15.09}\text{''}}\right)\;(= 29.(43\ldots))\) (29 – 30) \((BAC =)\sin^{-1}\left(\dfrac{9.2\sin\text{``}{118}\text{''}}{\text{``}{15.09\ldots}\text{''}}\right)\) or \(\cos^{-1}\left(\dfrac{8.4^2 + \text{``}{15.09}\text{''}^2 - 9.2^2}{2 \times 8.4 \times \text{``}{15.09}\text{''}}\right)\;(= 32.(56\ldots))\) (32 – 33) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Look for angles and lengths on their diagram Answer: 280 | A1 |
| (6) | |
| (6 marks) |
Notes
M1: A diagram showing 118 or use of 118 in further calculations
M1: a correct method to find length \(AC^2\)
M1: a correct method to find length \(AC\)
M1: dep on previous method marks
a correct statement of the sine rule or the cosine rule to find angle \(ACB\) or angle \(BAC\)
[numbers in inverted commas must come from correct working]
M1: a completely correct statement for angle \(ACB\) or angle \(BAC\)
[numbers in inverted commas must come from correct a correct method]
A1: allow 279 – 280
a correct answer from a scale drawing gains full marks but if answer is slightly inaccurate gains 0 marks