| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Bearings and navigation |
| Difficulty | Standard +0.3 This is a straightforward application of cosine rule for distance, sine rule for bearing, followed by basic arc length and angle conversion. All steps are standard procedures with clear setup from the diagram, requiring no novel insight—slightly easier than average due to the routine nature of each part. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(5.2^2 + 6.3^2 - 2 \times 5.2 \times 6.3 \times \cos 57°\) | M2 | M1 for recognisable attempt at cos rule |
| \(ST = 5.6\) or \(5.57\) cao | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{\sin T}{5.2} = \frac{\sin(\text{their }57)}{\text{their }ST}\) | M1 | Or \(\frac{\sin S}{6.3} = \ldots\) or cosine rule |
| \(T = 51\) to \(52\) or \(S = 71\) to \(72\) | A1 | |
| Bearing \(285 + \text{their }T\) or \(408 - \text{their }S\) | B1 | If outside 0 to 360, must be adjusted |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(5.2\theta\), \(24 \times \frac{26}{60}\) | B1B1 | Lost for all working in degrees |
| \(\theta = 1.98\) to \(2.02\) | B1 | Implied by 57.3 |
| \(\theta = \text{their } 2 \times \frac{180}{\pi}\) or \(114.6°...\) | M1 | |
| Bearing \(= 293\) to \(294\) cao | A1 |
## Question 4(i)(A):
| Answer | Mark | Guidance |
|--------|------|----------|
| $5.2^2 + 6.3^2 - 2 \times 5.2 \times 6.3 \times \cos 57°$ | M2 | M1 for recognisable attempt at cos rule |
| $ST = 5.6$ or $5.57$ cao | A1 | |
---
## Question 4(i)(B):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{\sin T}{5.2} = \frac{\sin(\text{their }57)}{\text{their }ST}$ | M1 | Or $\frac{\sin S}{6.3} = \ldots$ or cosine rule |
| $T = 51$ to $52$ or $S = 71$ to $72$ | A1 | |
| Bearing $285 + \text{their }T$ or $408 - \text{their }S$ | B1 | If outside 0 to 360, must be adjusted |
---
## Question 4(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $5.2\theta$, $24 \times \frac{26}{60}$ | B1B1 | Lost for all working in degrees |
| $\theta = 1.98$ to $2.02$ | B1 | Implied by 57.3 |
| $\theta = \text{their } 2 \times \frac{180}{\pi}$ or $114.6°...$ | M1 | |
| Bearing $= 293$ to $294$ cao | A1 | |
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\begin{enumerate}[label=(\roman*)]
\item \begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{cd507fa5-97b2-4edd-ae37-a58aea1de5ed-4_492_1018_256_567}
\captionsetup{labelformat=empty}
\caption{Fig. 10.1}
\end{center}
\end{figure}
At a certain time, ship S is 5.2 km from lighthouse L on a bearing of $048 ^ { \circ }$. At the same time, ship T is 6.3 km from L on a bearing of $105 ^ { \circ }$, as shown in Fig. 10.1.
For these positions, calculate\\
(A) the distance between ships S and T ,\\
(B) the bearing of S from T .
\item \begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{cd507fa5-97b2-4edd-ae37-a58aea1de5ed-4_430_698_1350_573}
\captionsetup{labelformat=empty}
\caption{Fig. 10.2}
\end{center}
\end{figure}
Not to scale
Ship S then travels at $24 \mathrm {~km} \mathrm {~h} { } ^ { 1 }$ anticlockwise along the arc of a circle, keeping 5.2 km from the lighthouse L, as shown in Fig. 10.2.
Find, in radians, the angle $\theta$ that the line LS has turned through in 26 minutes.\\
Hence find, in degrees, the bearing of ship S from the lighthouse at this time.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C2 Q4 [11]}}