| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Sequential triangle calculations (basic) |
| Difficulty | Moderate -0.8 Part (a) is a direct application of the cosine rule to find an angle given three sides (2 marks, routine). Part (b) requires calculating the area using the sine formula after finding the angle, then dividing by 1200 (3 marks, straightforward). This is a standard AS-level trigonometry question with no problem-solving insight required—just methodical 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 |
|---|---|---|
| 7(a) | Substitutes values correctly into | |
| cosine rule to find any angle | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| AWRT 46.3° | 1.1b | A1 |
| Subtotal | 2 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 7(b) | Uses 0.5 × b × c × sin A OE to | |
| find area | 3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Divides their area by 1200 | 1.1a | M1 |
| Obtains 21 | 3.2a | A1 |
| Subtotal | 3 | |
| Question 7 Total | 5 | |
| Q | Marking instructions | AO |
Question 7:
--- 7(a) ---
7(a) | Substitutes values correctly into
cosine rule to find any angle | 1.1a | M1 | 2252 = 2342 + 3102 –
2 × 234 × 310 × cosA
cos A = 0.690867
A = 46.3°
Obtains correct angle
AWRT 46.3° | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 7(b) ---
7(b) | Uses 0.5 × b × c × sin A OE to
find area | 3.1a | M1 | Area = 0.5 × 234 × 310 × sin A
= 26223
Number of sheep = 26223 ÷ 1200
= 21.85
Maximum number = 21
Divides their area by 1200 | 1.1a | M1
Obtains 21 | 3.2a | A1
Subtotal | 3
Question 7 Total | 5
Q | Marking instructions | AO | Marks | Typical solutionA triangular field of grass, $ABC$, has boundaries with lengths as follows:
$$AB = 234 \text{ m} \qquad BC = 225 \text{ m} \qquad AC = 310 \text{ m}$$
The field is shown in the diagram below.
\includegraphics{figure_7}
\begin{enumerate}[label=(\alph*)]
\item Find angle $A$
[2 marks]
\item Farmers calculate the number of sheep they can keep in a field, by allowing one sheep for every $1200 \text{ m}^2$ of grass.
Find the maximum number of sheep which can be kept in the field $ABC$
[3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2024 Q7 [5]}}