June 2023 Paper 3 Q2
2
| Scheme | Marks | AO |
|---|---|---|
| \(R = 5\) | B1 | 1.1 |
| \(R\cos\alpha = 3\) \(R\sin\alpha = 4\) \(\Rightarrow \tan\alpha = \dfrac{4}{3}\) | M1 | 1.1 |
| \(\alpha = 53.13\) | A1 | 1.1 |
| [3] |
Notes
B1: B0 for \(R = \pm 5, \sqrt{25}\) etc. unless replaced with 5
No working required for this mark. Ignore working
M1: M1 for \(\tan\alpha = k\) where \(k = \pm\frac{3}{4}, \pm\frac{4}{3}\) or equivalent e.g. \(\cos\alpha = \pm\dfrac{3}{R}\), \(\sin\alpha = \pm\dfrac{4}{R}\) with their value of \(R\) (but not just \(R\) and do not allow reciprocals for this mark). 53.1 (or better) with no working implies M1
SC If \(\cos\alpha = 3, \sin\alpha = 4 \Rightarrow \tan\alpha = \dfrac{4}{3}\) explicitly seen then this scores M1 A0 but do not penalise again in (b) (if correct answer seen)
A1: www awrt 53.13 (at least 4 sf required) so 53.1 (or 53) is A0 (but if an awrt 53.13 seen then isw if replaced with a less accurate value)
\(53.13010235\ldots\) - an answer in radians scores A0
53.13 from \(R\sin(x - \alpha)\) soi
| Scheme | Marks | AO |
|---|---|---|
| \(x = 53.13 + \arcsin\left(\dfrac{2}{5}\right)\) | M1 | 1.1 |
| \(x = 76.7\) | A1 | 1.1 |
| [2] |
Notes
M1: M1 for \(x = \alpha + \arcsin\left(\dfrac{2}{R}\right)\) or \(x - \alpha = \arcsin\left(\dfrac{2}{R}\right)\) with their \(R\) and \(\alpha\) substituted
SC B1 for 76.7 only (in the given range) from using an alternative method e.g. \(9\sin^2 x = (2 + 4\cos x)^2\)
A1: awrt 76.7 (at least 3 sf required) – ignore any answers given outside the range \(0 \lt x \lt 90\) but do not award this mark if any other values in this range are given – www but see SC in (a)
Correct answer with no working seen scores SC B1
Answer in radians scores A0