| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2008 |
| Session | January |
| Marks | 12 |
| 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 sine/cosine rules and arc length formula with clear diagrams and standard bearings work. Part (i) requires basic triangle calculations and bearing conversions, while part (ii) is a simple arc length plus chord calculation. All steps are routine with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| \(1033.86...[m]\) or ft \(650 +\) their \(BC\) | M2, A1, 1 | M1 for recognisable attempt at Cosine Rule to 3 sf or more; accept to 3 sf or more |
| Answer | Marks | Guidance |
|---|---|---|
| answer in range 306 to 307 | M1, A1, M1, A1 | Cosine Rule acceptable or Sine Rule to find C; or \(247 +\) their C |
| Answer | Marks | Guidance |
|---|---|---|
| answer in range 690 to 692 [m] | M2, B1, A1 | M1 for \(\frac{136}{360} \times 2\pi \times 120\) |
**iA**
$BC^2 = 348^2 + 302^2 - 2 \times 348 \times 302 \times \cos 72°$
$BC = 383.86...$
$1033.86...[m]$ or ft $650 +$ their $BC$ | M2, A1, 1 | M1 for recognisable attempt at Cosine Rule to 3 sf or more; accept to 3 sf or more | 4
**iB**
$\frac{\sin B}{302} = \frac{\sin 72°}{\text{their } BC}$
$B = 48.4...$
$355 -$ their B o.e.
answer in range 306 to 307 | M1, A1, M1, A1 | Cosine Rule acceptable or Sine Rule to find C; or $247 +$ their C | 4
**ii**
Arc length $PQ = \frac{224}{360} \times 2\pi \times 120$
o.e. or $469.1...$ to 3 sf or more
$QP = 222.5...$ to 3 sf or more
answer in range 690 to 692 [m] | M2, B1, A1 | M1 for $\frac{136}{360} \times 2\pi \times 120$ | 4
---11
\begin{enumerate}[label=(\roman*)]
\item The course for a yacht race is a triangle, as shown in Fig. 11.1. The yachts start at A , then travel to B , then to C and finally back to A .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{15872003-2e41-47e9-a5bd-34e533768f8a-4_661_869_404_680}
\captionsetup{labelformat=empty}
\caption{Fig. 11.1}
\end{center}
\end{figure}
(A) Calculate the total length of the course for this race.\\
(B) Given that the bearing of the first stage, AB , is $175 ^ { \circ }$, calculate the bearing of the second stage, BC.
\item Fig. 11.2 shows the course of another yacht race. The course follows the arc of a circle from P to $Q$, then a straight line back to $P$. The circle has radius 120 m and centre $O$; angle $P O Q = 136 ^ { \circ }$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{15872003-2e41-47e9-a5bd-34e533768f8a-4_709_821_1603_703}
\captionsetup{labelformat=empty}
\caption{Fig. 11.2}
\end{center}
\end{figure}
Calculate the total length of the course for this race.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C2 2008 Q11 [12]}}