| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Sequential triangle calculations (basic) |
| Difficulty | Moderate -0.8 This is a straightforward two-part question requiring direct application of the sine rule to find an angle, then the standard area formula. Both are routine calculations with no problem-solving or conceptual challenges—simpler than the average A-level question which typically requires more integration of techniques or multi-step reasoning. |
| 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 |
| \(\frac{10}{\sin\theta} = \frac{9}{\sin 54}\) | M1 | Sine rule, with \(\sin\theta\) being the only unknown |
| \(\sin\theta = \frac{10 \times \sin 54}{9}\) | m1 | Correct rearrangement to '\(\sin\theta = \ldots\)' or to '\(\theta = \sin^{-1}(\ldots)\)' |
| \(\sin\theta = 0.898(9\ldots)\), \(\theta = 64.01(48\ldots)\) | ||
| \(\theta = 64°\) {to nearest degree} | A1 | AG. m1 must have been awarded and must see at least 3sf value either for \(\sin\theta\) so that \(0.898 \leq \sin\theta \leq 0.8993\) or for \(\theta\) so that \(64.0 \leq \theta \leq 64.1\) as well as seeing '\(\theta\)(OE)= 64' |
| Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Angle \(C = 180 - (54+\theta) = 62\) {to 2sf} | B1 | \(C = 62\). AWRT 62. PI if '\(C = 180-(54+\theta)\)' and accurate later work |
| Area \(= \frac{1}{2} \times 10 \times 9 \sin 62\) | M1 | OE. Ft c's value for \(C\) (\(C \neq 54\), \(C \neq \theta\)) |
| \(= 39.73\ldots = 40\{\text{cm}^2 \text{ to nearest sq cm}\}\) | A1 | If not 40 condone a value 39.7 to 39.8 inclusive |
| Total: 3 marks |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{10}{\sin\theta} = \frac{9}{\sin 54}$ | M1 | Sine rule, with $\sin\theta$ being the only unknown |
| $\sin\theta = \frac{10 \times \sin 54}{9}$ | m1 | Correct rearrangement to '$\sin\theta = \ldots$' or to '$\theta = \sin^{-1}(\ldots)$' |
| $\sin\theta = 0.898(9\ldots)$, $\theta = 64.01(48\ldots)$ | | |
| $\theta = 64°$ {to nearest degree} | A1 | AG. m1 must have been awarded and must see at least 3sf value either for $\sin\theta$ so that $0.898 \leq \sin\theta \leq 0.8993$ or for $\theta$ so that $64.0 \leq \theta \leq 64.1$ as well as seeing '$\theta$(OE)= 64' |
| **Total: 3 marks** | | |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Angle $C = 180 - (54+\theta) = 62$ {to 2sf} | B1 | $C = 62$. AWRT 62. PI if '$C = 180-(54+\theta)$' and accurate later work |
| Area $= \frac{1}{2} \times 10 \times 9 \sin 62$ | M1 | OE. Ft c's value for $C$ ($C \neq 54$, $C \neq \theta$) |
| $= 39.73\ldots = 40\{\text{cm}^2 \text{ to nearest sq cm}\}$ | A1 | If not 40 condone a value 39.7 to 39.8 inclusive |
| **Total: 3 marks** | | |
---1 The triangle $A B C$, shown in the diagram, is such that $A C = 9 \mathrm {~cm} , B C = 10 \mathrm {~cm}$, angle $A B C = 54 ^ { \circ }$ and the acute angle $B A C = \theta$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\theta = 64 ^ { \circ }$, correct to the nearest degree.
\item Calculate the area of triangle $A B C$, giving your answer to the nearest square centimetre.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2011 Q1 [6]}}