| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Sequential triangle calculations (basic) |
| Difficulty | Easy -1.2 This is a straightforward two-part question requiring direct application of the sine rule (part a) and triangle area formula (part b). Both are standard C2 techniques with no problem-solving insight needed—students simply substitute given values into formulas. The 'show that' format in part (a) provides the answer, making it easier than an open calculation. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{\sin\theta}{6} = \frac{\sin 150°}{15}\) | M1 | Use of sine rule |
| \(\sin\theta = \frac{6\sin 150°}{15} = \frac{6 \times 0.5}{15} = 0.2\) | A1 | Correct expression |
| \(\theta = 11.537...° \approx 11.5°\) | A1 | Conclusion stated to nearest \(0.1°\) |
| Answer | Marks | Guidance |
|---|---|---|
| Angle \(ABC = 180° - 150° - 11.5° = 18.5°\) | M1 | Finding third angle |
| Area \(= \frac{1}{2} \times 6 \times 15 \times \sin 18.5°\) | M1 | Use of \(\frac{1}{2}ab\sin C\) |
| \(= 28.4 \text{ cm}^2\) | A1 | 3 significant figures |
# Question 3:
## Part (a)
| $\frac{\sin\theta}{6} = \frac{\sin 150°}{15}$ | M1 | Use of sine rule |
|---|---|---|
| $\sin\theta = \frac{6\sin 150°}{15} = \frac{6 \times 0.5}{15} = 0.2$ | A1 | Correct expression |
| $\theta = 11.537...° \approx 11.5°$ | A1 | Conclusion stated to nearest $0.1°$ |
## Part (b)
| Angle $ABC = 180° - 150° - 11.5° = 18.5°$ | M1 | Finding third angle |
|---|---|---|
| Area $= \frac{1}{2} \times 6 \times 15 \times \sin 18.5°$ | M1 | Use of $\frac{1}{2}ab\sin C$ |
| $= 28.4 \text{ cm}^2$ | A1 | 3 significant figures |
---3 The triangle $A B C$, shown in the diagram, is such that $A B = 6 \mathrm {~cm} , B C = 15 \mathrm {~cm}$, angle $B A C = 150 ^ { \circ }$ and angle $A C B = \theta$.\\
\includegraphics[max width=\textwidth, alt={}, center]{f9a7a4dd-f7fd-4135-8872-2c1270d46a14-4_376_867_406_584}
\begin{enumerate}[label=(\alph*)]
\item Show that $\theta = 11.5 ^ { \circ }$, correct to the nearest $0.1 ^ { \circ }$.
\item Calculate the area of triangle $A B C$, giving your answer in $\mathrm { cm } ^ { 2 }$ to three significant figures.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2010 Q3 [6]}}