| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2012 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Given area find angle/side |
| Difficulty | Standard +0.3 This is a straightforward multi-part question using standard area formula and cosine rule. Part (a) is a 'show that' using area = ½ab sin C with simple arithmetic. Parts (b) and (c) apply cosine and sine rules directly with no conceptual challenges—slightly easier than average due to the scaffolded structure and routine application of formulas. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Area \(= \frac{1}{2}\times10\times AC\sin150\) | M1 | \(\frac{1}{2}\times10\times AC\sin150\) |
| \(40 = 2.5AC\) so \(AC = 16\) (m) | A1 (Total: 2) | AG Be convinced |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(BC^2 = 10^2+16^2-2\times10\times16\times\cos150 = 100+256+277.128\ldots\) | M1, m1 | RHS of cosine rule used; correct order of evaluation |
| \(BC = \sqrt{633.128\ldots} = 25.162\ldots = 25.16\)m | A1 (Total: 3) | AWRT 25.16 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{10}{\sin C} = \frac{BC}{\sin150}\) | M1 | A correct equation using sine rule or cosine rule or area formula for either \(B\) or \(C\). Subst of \(BC\) or \(AC\) not required for this M |
| \(\sin C = \frac{10\sin150}{\text{"25.16"}}\) \((=0.1987..)\) | m1 | Correct rearrangement to either \(\sin C\) or \(\cos C\) or \(\sin B\) or \(\cos B\) equal to numerical expression ft on c's numerical value for \(BC\) |
| Smallest angle \((C=)\) \(11.5°\) to 1dp | A1 (Total: 3) | Accept a value 11.4 to 11.5 inclusive |
## Question 4:
**Part (a):**
| Working | Mark | Guidance |
|---------|------|----------|
| Area $= \frac{1}{2}\times10\times AC\sin150$ | M1 | $\frac{1}{2}\times10\times AC\sin150$ |
| $40 = 2.5AC$ so $AC = 16$ (m) | A1 (Total: 2) | AG Be convinced |
**Part (b):**
| Working | Mark | Guidance |
|---------|------|----------|
| $BC^2 = 10^2+16^2-2\times10\times16\times\cos150 = 100+256+277.128\ldots$ | M1, m1 | RHS of cosine rule used; correct order of evaluation |
| $BC = \sqrt{633.128\ldots} = 25.162\ldots = 25.16$m | A1 (Total: 3) | AWRT 25.16 |
**Part (c):**
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{10}{\sin C} = \frac{BC}{\sin150}$ | M1 | A correct equation using sine rule or cosine rule or area formula for either $B$ or $C$. Subst of $BC$ or $AC$ not required for this M |
| $\sin C = \frac{10\sin150}{\text{"25.16"}}$ $(=0.1987..)$ | m1 | Correct rearrangement to either $\sin C$ or $\cos C$ or $\sin B$ or $\cos B$ equal to numerical expression ft on c's numerical value for $BC$ |
| Smallest angle $(C=)$ $11.5°$ to 1dp | A1 (Total: 3) | Accept a value 11.4 to 11.5 inclusive |
---4 The triangle $A B C$, shown in the diagram, is such that $A B$ is 10 metres and angle $B A C$ is $150 ^ { \circ }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{02e5dfac-18d7-480d-ac23-dfd2ca348cba-3_323_746_406_648}
The area of triangle $A B C$ is $40 \mathrm {~m} ^ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that the length of $A C$ is 16 metres.
\item Calculate the length of $B C$, giving your answer, in metres, to two decimal places.
\item Calculate the smallest angle of triangle $A B C$, giving your answer to the nearest $0.1 ^ { \circ }$.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2012 Q4 [8]}}