| Exam Board | Edexcel |
|---|---|
| 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 | Quadrilateral with diagonal |
| Difficulty | Standard +0.3 This is a straightforward application of the sine rule in two triangles sharing a diagonal. Part (a) requires one sine rule calculation to find x, and part (b) requires using the cosine rule after finding the angle at D. The question is slightly easier than average because it's a standard two-part question with clear setup, routine trigonometric rules, and no conceptual obstacles—students just need to identify which rule to apply and execute calculations carefully. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{x}{\sin 30°} = \frac{x+3}{\sin 140°}\) | M1 | Recognises need for sine rule with sides in correct positions to form equation in \(x\) |
| \(2x\sin 140° - x = 3\) | dM1 | Attempts to rearrange collecting \(x\) terms on one side; dependent on previous M1; condone sign slips only |
| \(x = 10.5\) | A1* | Must have achieved correct expression for \(x\) or \(x(2\sin 140° - 1) = 3\); do not allow verification attempts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(AB^2 = 8.5^2 + 13.5^2 - 2 \times 8.5 \times 13.5 \times \cos 150°\) | M1 | Applies cosine rule correctly using \(150°\) for angle \(ADB\); substitutes \(x = 10.5\) |
| \(AB = \text{awrt } 21.3 \text{ (cm)}\) | A1 | Units not required but if given must be correct |
## Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{x}{\sin 30°} = \frac{x+3}{\sin 140°}$ | M1 | Recognises need for sine rule with sides in correct positions to form equation in $x$ |
| $2x\sin 140° - x = 3$ | dM1 | Attempts to rearrange collecting $x$ terms on one side; dependent on previous M1; condone sign slips only |
| $x = 10.5$ | A1* | Must have achieved correct expression for $x$ or $x(2\sin 140° - 1) = 3$; do not allow verification attempts |
## Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $AB^2 = 8.5^2 + 13.5^2 - 2 \times 8.5 \times 13.5 \times \cos 150°$ | M1 | Applies cosine rule correctly using $150°$ for angle $ADB$; substitutes $x = 10.5$ |
| $AB = \text{awrt } 21.3 \text{ (cm)}$ | A1 | Units not required but if given must be correct |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{23689deb-7eed-4022-848f-1278231a4056-10_547_1475_306_294}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of triangle $A B D$ and triangle $B C D$\\
Given that
\begin{itemize}
\item $A D C$ is a straight line
\item $B D = ( x + 3 ) \mathrm { cm }$
\item $B C = x \mathrm {~cm}$
\item angle $B D C = 30 ^ { \circ }$
\item angle $B C D = 140 ^ { \circ }$
\begin{enumerate}[label=(\alph*)]
\item show that $x = 10.5$ correct to 3 significant figures.
\end{itemize}
Given also that $A D = ( x - 2 ) \mathrm { cm }$
\item find the length of $A B$, giving your answer to 3 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 1 2024 Q4 [5]}}