| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2008 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Perpendicular from vertex |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing standard applications of the cosine rule, area formula (½ab sin C), and using area to find a perpendicular height. All parts follow routine procedures with no problem-solving insight required, making it easier than average but not trivial due to the 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 |
|---|---|---|
| (a) \([BC^2] = 7.6^2 + 8.3^2 - 2 \times 7.6 \times 8.3\cos 65\ldots = 57.76 + 68.89 - 53.3175\ldots\) | M1, m1 (3 marks) | RHS of cosine rule used; Correct order of evaluation |
| \(BC = \sqrt{73.33\ldots} = 8.563\ldots (= 8.56\) m) | A1 | AG; must see \(\sqrt{73.33\ldots}\) or > 3sf value |
| (b) Area triangle \(= \frac{1}{2} \times 7.6 \times 8.3 \times \sin 65 = 28.58\ldots = 28.6\) (m²) | M1, A1 (2 marks) | Use of \(\frac{1}{2}bc\sin A\) OE; Condone > 3sf |
| (c) Area of triangle \(= 0.5 \times BC \times AD\); \(AD = \frac{\text{Ans (b)} \times 10.5 \times \text{Ans (a)}}{...}\) | M1, m1 (3 marks) | Or valid method to find \(\sin B\) or \(\sin C\); Or \(AD = 7.6\sin B\); Or \(AD = 8.3\sin C\) |
| \(AD = 6.67\ldots = 6.7\) (m) | A1 | If not 6.7 accept 6.65 to 6.69 inclusive. |
**(a)** $[BC^2] = 7.6^2 + 8.3^2 - 2 \times 7.6 \times 8.3\cos 65\ldots = 57.76 + 68.89 - 53.3175\ldots$ | M1, m1 (3 marks) | RHS of cosine rule used; Correct order of evaluation
$BC = \sqrt{73.33\ldots} = 8.563\ldots (= 8.56$ m) | A1 | AG; must see $\sqrt{73.33\ldots}$ or > 3sf value
**(b)** Area triangle $= \frac{1}{2} \times 7.6 \times 8.3 \times \sin 65 = 28.58\ldots = 28.6$ (m²) | M1, A1 (2 marks) | Use of $\frac{1}{2}bc\sin A$ OE; Condone > 3sf
**(c)** Area of triangle $= 0.5 \times BC \times AD$; $AD = \frac{\text{Ans (b)} \times 10.5 \times \text{Ans (a)}}{...}$ | M1, m1 (3 marks) | Or valid method to find $\sin B$ or $\sin C$; Or $AD = 7.6\sin B$; Or $AD = 8.3\sin C$
$AD = 6.67\ldots = 6.7$ (m) | A1 | If not 6.7 accept 6.65 to 6.69 inclusive.
**Total for Q4: 8 marks**
---4 The diagram shows a triangle $A B C$.\\
\includegraphics[max width=\textwidth, alt={}, center]{a2525df8-dbd0-4b69-b6bb-f8ef6f96f7dc-3_394_522_1062_751}
The size of angle $B A C$ is $65 ^ { \circ }$, and the lengths of $A B$ and $A C$ are 7.6 m and 8.3 m respectively.
\begin{enumerate}[label=(\alph*)]
\item Show that the length of $B C$ is 8.56 m , correct to three significant figures.
\item Calculate the area of triangle $A B C$, giving your answer in $\mathrm { m } ^ { 2 }$ to three significant figures.
\item The perpendicular from $A$ to $B C$ meets $B C$ at the point $D$.
Calculate the length of $A D$, giving your answer to the nearest 0.1 m .
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2008 Q4 [8]}}