June 2023 Paper 1 Q1
1 In the triangle \(ABC\), the length \(AB = 6\,\text{cm}\), the length \(AC = 15\,\text{cm}\) and the angle \(BAC = 30^\circ\).
\(D\) is the point on \(AC\) such that the length \(BD = 4\,\text{cm}\).
| Scheme | Marks | AO |
|---|---|---|
| \(BC^2 = 6^2 + 15^2 - 2 \times 6 \times 15 \times \cos 30^\circ\) | M1 | 1.1a |
| \(BC = 10.3\) cm | A1 | 1.1 |
| [2] |
Notes
M1: Attempt use of cosine rule
Allow either omission of 2, or + not –, but no other errors
Allow other fully complete methods, such as basic trigonometry, possibly combined with Pythagoras
A1: Obtain 10.3cm, or better
If > 3sf then allow 10.25, or answers that round to 10.25
Condone no units
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\sin 30}{4} = \dfrac{\sin D}{6}\) | M1 | 1.1 |
| \(D = 48.6^\circ\) | A1 | 1.1 |
| or \(D = 131^\circ\) | A1FT | 3.1a |
| [3] |
Notes
M1: Attempt use of sine rule
Correct equation seen, with fractions either way up
Could also be implied by method eg \(\sin^{-1}(0.75)\) is M1, but just 0.75 is M0
Allow other fully complete methods
A1: Obtain \(D = 48.6^\circ\), or better
\(D = 48.590377\ldots\)
Allow \(D = 0.848\) radians
A1FT: Obtain \(D = 131^\circ\), or better
FT their first angle as long as \(\lt 150^\circ\)
\(D = 131.409622\ldots\)
A0 if additional angles given as well
Allow \(D = 2.29\) radians (could be FT on incorrect acute angle in radians, as long as \(D \lt 2.618\))