| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Algebraic side lengths |
| Difficulty | Standard +0.3 This is a straightforward application of the cosine rule to create a quadratic equation, followed by routine solving and using the sine rule. The algebraic manipulation is standard AS-level work with clear signposting ('Show that'), making it slightly easier than average but still requiring multiple connected steps. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.05b Sine and cosine rules: including ambiguous case |
| Answer | Marks | Guidance |
|---|---|---|
| \((3x+10)^2 = (x+2)^2 + (7x)^2 - 2(x+2)(7x)\cos 60°\) | M1 | Recognises need to apply cosine rule with correct positions; condone invisible brackets and slips on \(3x+10\) as \(3x-10\) |
| Uses \(\cos 60° = \frac{1}{2}\), expands brackets and proceeds to a 3 term quadratic equation | dM1 | Dependent on M1; must proceed to 3TQ |
| \(17x^2 - 35x - 48 = 0\)* | A1* | Correct quadratic with \(= 0\), no errors in main body; condone recovery of invisible brackets |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 3\) | B1 | Selects appropriate value \(x = 3\) only; other root must be rejected or \(x=3\) used in (b) |
| Answer | Marks | Guidance |
|---|---|---|
| or \(5^2 = 21^2 + 19^2 - 2\times19\times21\cos ACB \Rightarrow \cos ACB = \frac{37}{38}\) | M1 | Using their \(x\) value; applying sine rule correctly to get \(\sin ACB\) or cosine rule to get \(\cos ACB\); condone slips calculating \(AB\), \(BC\), \(AC\) |
| \(\theta =\) awrt \(13.2\) | A1 | awrt 13.2 |
## Question 4(a)(i):
$(3x+10)^2 = (x+2)^2 + (7x)^2 - 2(x+2)(7x)\cos 60°$ | M1 | Recognises need to apply cosine rule with correct positions; condone invisible brackets and slips on $3x+10$ as $3x-10$
Uses $\cos 60° = \frac{1}{2}$, expands brackets and proceeds to a 3 term quadratic equation | dM1 | Dependent on M1; must proceed to 3TQ
$17x^2 - 35x - 48 = 0$* | A1* | Correct quadratic with $= 0$, no errors in main body; condone recovery of invisible brackets
**Total: 3 marks**
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## Question 4(a)(ii):
$x = 3$ | B1 | Selects appropriate value $x = 3$ only; other root must be rejected or $x=3$ used in (b)
**Total: 1 mark**
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## Question 4(b):
$\frac{5}{\sin ACB} = \frac{19}{\sin 60°} \Rightarrow \sin ACB = \frac{5\sqrt{3}}{38}$
or $5^2 = 21^2 + 19^2 - 2\times19\times21\cos ACB \Rightarrow \cos ACB = \frac{37}{38}$ | M1 | Using their $x$ value; applying sine rule correctly to get $\sin ACB$ or cosine rule to get $\cos ACB$; condone slips calculating $AB$, $BC$, $AC$
$\theta =$ awrt $13.2$ | A1 | awrt 13.2
**Total: 2 marks**
---4.
Figure 1
Figure 1 shows a sketch of triangle $A B C$ with $A B = ( x + 2 ) \mathrm { cm } , B C = ( 3 x + 10 ) \mathrm { cm }$, $A C = 7 x \mathrm {~cm}$, angle $B A C = 60 ^ { \circ }$ and angle $A C B = \theta ^ { \circ }$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that $17 x ^ { 2 } - 35 x - 48 = 0$
\item Hence find the value of $x$.
\end{enumerate}\item Hence find the value of $\theta$ giving your answer to one decimal place.
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 1 2022 Q4 [6]}}