| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Exact trigonometric values |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question requiring direct application of the cosine rule, the Pythagorean identity sin²θ + cos²θ = 1, and the area formula (1/2)ab sin C. All steps are routine with clear signposting ('show that' and 'hence'), requiring only standard technique with no problem-solving or insight. Easier than average due to its highly structured nature and use of exact values. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C)1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(6^2 = 4^2 + 5^2 - 2(4)(5)\cos\theta\) | M1 |
| \(\cos\theta = \frac{4^2 + 5^2 - 6^2}{2(4)(5)}\) | m1 | Rearrangement |
| \(\cos\theta = \frac{5}{40} = \frac{1}{8}\) | A1 | CSO AG (be convinced) |
| Total: 3 | ||
| (b) | \(\cos^2\theta + \sin^2\theta = 1\) | M1 |
| \(\sin^2\theta = \frac{63}{64}\) | A1 | Or better |
| \(\sin\theta = \frac{\sqrt{63}}{8} = \frac{\sqrt{9 \times 7}}{8} = \frac{3\sqrt{7}}{8}\) | A1 | AG (be convinced) |
| Total: 3 | ||
| (c) | Area of triangle \(= 0.5 \times 4 \times 5 \times \sin\theta\) | M1 |
| \(..... = \frac{30\sqrt{7}}{8} \text{ cm}^2\) | A1 | OE (Condone 9.92) |
| Total: 8 |
(a) | $6^2 = 4^2 + 5^2 - 2(4)(5)\cos\theta$ | M1 | Use of the cosine rule |
| $\cos\theta = \frac{4^2 + 5^2 - 6^2}{2(4)(5)}$ | m1 | Rearrangement |
| $\cos\theta = \frac{5}{40} = \frac{1}{8}$ | A1 | CSO AG (be convinced) |
| | **Total: 3** | |
(b) | $\cos^2\theta + \sin^2\theta = 1$ | M1 | Stated or used (PI) |
| $\sin^2\theta = \frac{63}{64}$ | A1 | Or better |
| $\sin\theta = \frac{\sqrt{63}}{8} = \frac{\sqrt{9 \times 7}}{8} = \frac{3\sqrt{7}}{8}$ | A1 | AG (be convinced) |
| | **Total: 3** | |
(c) | Area of triangle $= 0.5 \times 4 \times 5 \times \sin\theta$ | M1 | |
| $..... = \frac{30\sqrt{7}}{8} \text{ cm}^2$ | A1 | OE (Condone 9.92) |
| | **Total: 8** | |
---4 The triangle $A B C$, shown in the diagram, is such that $B C = 6 \mathrm {~cm} , A C = 5 \mathrm {~cm}$ and $A B = 4 \mathrm {~cm}$. The angle $B A C$ is $\theta$.\\
\includegraphics[max width=\textwidth, alt={}, center]{c16d94a6-52f2-4bf3-acee-0b227ae55a1a-3_442_652_452_678}
\begin{enumerate}[label=(\alph*)]
\item Use the cosine rule to show that $\cos \theta = \frac { 1 } { 8 }$.
\item Hence use a trigonometrical identity to show that $\sin \theta = \frac { 3 \sqrt { 7 } } { 8 }$.
\item Hence find the area of the triangle $A B C$.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2007 Q4 [8]}}