| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Given area find angle/side |
| Difficulty | Moderate -0.3 This is a straightforward application of the triangle area formula (½ab sin C) followed by the cosine rule. Part (a) requires solving a simple equation involving surds, and part (b) is direct substitution into the cosine rule. Both parts are routine techniques with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Uses \(18\sqrt{3} = \frac{1}{2} \times 2x \times 3x \times \sin 60°\) | M1 | 1.1a – Attempts formula \(A = \frac{1}{2}ab\sin C\); writing \(18\sqrt{3} = \frac{1}{2} \times 5x \times \sin 60°\) without prior correct line is M0 |
| Sight of \(\sin 60° = \frac{\sqrt{3}}{2}\) and proceeds to \(x^2 = k\) | M1 | 1.1b – Or awrt 0.866; may be awarded from correct formula or \(A = ab\sin C\) |
| \(x = \sqrt{12} = 2\sqrt{3}\) * | A1* | 2.1 – Given answer; look for \(x^2 = 12\) or \(x^2 = 4 \times 3\); cannot be scored via decimals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Uses \(BC^2 = (6\sqrt{3})^2 + (4\sqrt{3})^2 - 2 \times 6\sqrt{3} \times 4\sqrt{3} \times \cos 60°\) | M1 | 1.1b – Cosine rule with sides in correct position; can be scored from \(BC^2 = (3x)^2 + (2x)^2 - 2 \times 3x \times 2x \times \cos 60°\) with later substitution; condone slips on squaring |
| \(BC^2 = 84\) | A1 | 1.1b – Accept \(BC^2 = 7 \times 12\), \(BC = \sqrt{84}\); if surds replaced by decimals, A1 for \(BC^2 =\) awrt 84.0 |
| \(BC = 2\sqrt{21}\) (cm) | A1 | 1.1b – Condone other variables e.g. \(x = 2\sqrt{21}\); cannot be scored via decimals |
## Question 6:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Uses $18\sqrt{3} = \frac{1}{2} \times 2x \times 3x \times \sin 60°$ | M1 | 1.1a – Attempts formula $A = \frac{1}{2}ab\sin C$; writing $18\sqrt{3} = \frac{1}{2} \times 5x \times \sin 60°$ without prior correct line is M0 |
| Sight of $\sin 60° = \frac{\sqrt{3}}{2}$ and proceeds to $x^2 = k$ | M1 | 1.1b – Or awrt 0.866; may be awarded from correct formula or $A = ab\sin C$ |
| $x = \sqrt{12} = 2\sqrt{3}$ * | A1* | 2.1 – Given answer; look for $x^2 = 12$ or $x^2 = 4 \times 3$; cannot be scored via decimals |
**(3 marks)**
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Uses $BC^2 = (6\sqrt{3})^2 + (4\sqrt{3})^2 - 2 \times 6\sqrt{3} \times 4\sqrt{3} \times \cos 60°$ | M1 | 1.1b – Cosine rule with sides in correct position; can be scored from $BC^2 = (3x)^2 + (2x)^2 - 2 \times 3x \times 2x \times \cos 60°$ with later substitution; condone slips on squaring |
| $BC^2 = 84$ | A1 | 1.1b – Accept $BC^2 = 7 \times 12$, $BC = \sqrt{84}$; if surds replaced by decimals, A1 for $BC^2 =$ awrt 84.0 |
| $BC = 2\sqrt{21}$ (cm) | A1 | 1.1b – Condone other variables e.g. $x = 2\sqrt{21}$; cannot be scored via decimals |
**(3 marks)**
---6.
Figure 1
Figure 1 shows a sketch of a triangle $A B C$ with $A B = 3 x \mathrm {~cm} , A C = 2 x \mathrm {~cm}$ and angle $C A B = 60 ^ { \circ }$
Given that the area of triangle $A B C$ is $18 \sqrt { 3 } \mathrm {~cm} ^ { 2 }$
\begin{enumerate}[label=(\alph*)]
\item show that $x = 2 \sqrt { 3 }$
\item Hence find the exact length of BC, giving your answer as a simplified surd.
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 1 2019 Q6 [6]}}