| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Algebraic side lengths |
| Difficulty | Standard +0.8 This question requires understanding that the largest angle is opposite the longest side, applying the cosine rule with algebraic side lengths (requiring careful expansion and simplification), then using the circle geometry property that angle in a semicircle is 90° to set cos A = 0 and solve. The algebraic manipulation in part (ii) and the geometric insight in part (b) elevate this above a routine sine/cosine rule application. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The largest angle must be opposite the largest side which is \((m+n)\) | E1 | AO 2.4 — explains that \((m+n)\) is the largest side (opposite the largest angle) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((m+n)^2 = m^2 + (m-n)^2 - 2m(m-n)\cos A\) | M1 | AO 1.1a — substitutes \(a\), \(b\), \(c\) into cosine rule |
| \(\cos A = \frac{m^2 + (m-n)^2 - (m+n)^2}{2m(m-n)}\) | M1 | AO 1.1a — makes \(\cos A\) the subject |
| \(\cos A = \frac{m^2 - 4mn}{2m(m-n)} = \frac{m-4n}{2(m-n)}\) | R1 | AO 2.1 — completes reasoned argument to obtain given result AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A\) must be \(90°\), so \(\cos A = 0\) | B1 | AO 2.2a — deduces \(A\) is \(90°\); accept marked on diagram; PI by awarding M1 |
| \(\frac{m-4n}{2(m-n)} = 0 \Rightarrow m = 4n\) | M1 | AO 3.1a — substitutes \(\cos A = 0\) in part (a) equation; PI by correct value for \(n\) |
| \(n = 2\) | A1 | AO 1.1b |
## Question 8(a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The largest angle must be opposite the largest side which is $(m+n)$ | E1 | AO 2.4 — explains that $(m+n)$ is the largest side (opposite the largest angle) |
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## Question 8(a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(m+n)^2 = m^2 + (m-n)^2 - 2m(m-n)\cos A$ | M1 | AO 1.1a — substitutes $a$, $b$, $c$ into cosine rule |
| $\cos A = \frac{m^2 + (m-n)^2 - (m+n)^2}{2m(m-n)}$ | M1 | AO 1.1a — makes $\cos A$ the subject |
| $\cos A = \frac{m^2 - 4mn}{2m(m-n)} = \frac{m-4n}{2(m-n)}$ | R1 | AO 2.1 — completes reasoned argument to obtain given result AG |
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## Question 8(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A$ must be $90°$, so $\cos A = 0$ | B1 | AO 2.2a — deduces $A$ is $90°$; accept marked on diagram; PI by awarding M1 |
| $\frac{m-4n}{2(m-n)} = 0 \Rightarrow m = 4n$ | M1 | AO 3.1a — substitutes $\cos A = 0$ in part (a) equation; PI by correct value for $n$ |
| $n = 2$ | A1 | AO 1.1b |8 Triangle $A B C$ has sides of length $( m - n ) , m$ and $( m + n )$ where $0 < 2 n < m$
Angle $A$ is the largest angle in the triangle.\\
8
\begin{enumerate}[label=(\alph*)]
\item (i) Explain why angle $A$ must be opposite the side of length $( m + n )$.
8 (a) (ii) Using the cosine rule, show that $\cos A = \frac { m - 4 n } { 2 ( m - n ) }$\\
8
\item You are given that $B C$ is the diameter of a circle, and $A$ lies on the circumference of the circle. The value of $m$ is 8
Calculate the value of $n$.
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 2 2022 Q8 [7]}}