| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2021 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Basic cosine rule application |
| Difficulty | Easy -1.2 This is a straightforward application of the cosine rule requiring direct substitution into the formula, followed by a trivial observation to state AD. It's below average difficulty as it's a routine AS-level calculation with no problem-solving or insight required beyond formula recall. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case |
| Answer | Marks | Guidance |
|---|---|---|
| \(7^2 = 8^2 + x^2 - 2 \times 8 \times x \times \cos 60\) | B1, M1 | Uses cosine rule; forms expression for AC |
| \(x^2 - 8x + 15 = 0\), so \(x = 3\) or \(5\) | — | — |
| \(AC = 5\) cm | R1 | Deduces \(AC = 5\) cm (AWRT); condone missing/incorrect units |
| Answer | Marks | Guidance |
|---|---|---|
| \(AD = 3\) cm | B1 | Condone missing/incorrect units |
## Question 5(a):
$7^2 = 8^2 + x^2 - 2 \times 8 \times x \times \cos 60$ | B1, M1 | Uses cosine rule; forms expression for AC |
$x^2 - 8x + 15 = 0$, so $x = 3$ or $5$ | — | — |
$AC = 5$ cm | R1 | Deduces $AC = 5$ cm (AWRT); condone missing/incorrect units |
## Question 5(b):
$AD = 3$ cm | B1 | Condone missing/incorrect units |
---5
\begin{enumerate}[label=(\alph*)]
\item Using the cosine rule, find the length of $A C$.\\
5
\item Hence, state the length of $A D$.
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 2 2021 Q5 [4]}}