| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | January |
| 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 two-part application of standard formulas: using area = ½ab sin C to find an obtuse angle, then applying the cosine rule. Both are direct substitutions with minimal problem-solving required, making it slightly easier than average but not trivial due to the obtuse angle specification and multi-step nature. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\frac{1}{2}\times 5\times 6\sin C = 12.5\) | M1 | Area \(= \frac{1}{2}\times 5\times 6\sin C\) |
| \(\sin C = 0.833(3\ldots)\) | A1 | AWRT 0.83 or 5/6 OE; PI by seeing 56 or better |
| (\(C\) is obtuse) \(C = 123.6°\) | A1 | AWRT 123.6 |
| Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(AB^2 = 5^2+6^2-2\times 5\times 6\cos C\) | M1 | RHS of cosine rule used |
| \(= 61-60\times(-0.553\ldots) = 94.1(66\ldots)\) | m1 | Correct ft evaluation to at least 2sf of \(AB^2\) or \(AB\) using c's value of \(C\) |
| \(AB = 9.7\) (cm to 2sf) | A1 | If not 9.7 accept AWRT 9.70 or AWRT 9.71 |
| Total: 3 marks |
# Question 3:
## Part (a)
| Working | Marks | Guidance |
|---------|-------|----------|
| $\frac{1}{2}\times 5\times 6\sin C = 12.5$ | M1 | Area $= \frac{1}{2}\times 5\times 6\sin C$ |
| $\sin C = 0.833(3\ldots)$ | A1 | AWRT 0.83 or 5/6 OE; PI by seeing 56 or better |
| ($C$ is obtuse) $C = 123.6°$ | A1 | AWRT 123.6 |
| **Total: 3 marks** | | |
## Part (b)
| Working | Marks | Guidance |
|---------|-------|----------|
| $AB^2 = 5^2+6^2-2\times 5\times 6\cos C$ | M1 | RHS of cosine rule used |
| $= 61-60\times(-0.553\ldots) = 94.1(66\ldots)$ | m1 | Correct ft evaluation to at least 2sf of $AB^2$ or $AB$ using c's value of $C$ |
| $AB = 9.7$ (cm to 2sf) | A1 | If not 9.7 accept AWRT 9.70 or AWRT 9.71 |
| **Total: 3 marks** | | |
---3 The diagram shows a triangle $A B C$.\\
\includegraphics[max width=\textwidth, alt={}, center]{bfe96138-9587-4efb-95c5-84c4d5eadfbe-3_273_622_356_708}
The lengths of $A C$ and $B C$ are 5 cm and 6 cm respectively.\\
The area of triangle $A B C$ is $12.5 \mathrm {~cm} ^ { 2 }$, and angle $A C B$ is obtuse.
\begin{enumerate}[label=(\alph*)]
\item Find the size of angle $A C B$, giving your answer to the nearest $0.1 ^ { \circ }$.
\item Find the length of $A B$, giving your answer to two significant figures.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2013 Q3 [6]}}