| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Bearings and navigation |
| Difficulty | Moderate -0.3 This is a multi-part question covering standard C2 topics (sine/cosine rules, bearings, sector areas, trapezium rule). Part (a) requires cosine rule then sine rule for bearings - routine application. Part (b) involves standard sector area and showing a given result using trigonometry. The trapezium rule application is straightforward. While multi-step, each component is a textbook exercise requiring recall and standard procedures rather than problem-solving insight, making it slightly easier than average. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C)1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| \(10.6^2 + 9.2^2 - 2 \times 10.6 \times 9.2 \times \cos 68°\) | M1 | |
| \(QR = 11.1(3...)\) | A1 | |
| \(\frac{\sin 68}{\text{their } QR} = \frac{\sin Q}{9.2}\) or \(\frac{\sin R}{10.6}\) | M1 | Or correct use of Cosine Rule |
| \(Q = 50.01...°\) or \(R = 61.98...°\) | A1 | 2 s.f. or better |
| Bearing \(= 174.9\) to \(175°\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \((A)\ \frac{1}{2} \times 80^2 \times \frac{2\pi}{3}\) | M1 | |
| \(= \frac{6400\pi}{3}\) | A1 | \(6702.(...)\) to 2 s.f. or more |
| Answer | Marks | Guidance |
|---|---|---|
| \(DC = 80\sin\!\left(\frac{\pi}{3}\right) = 80\frac{\sqrt{3}}{2}\) | B1 | Both steps required |
| Area \(= \frac{1}{2} \times \text{their } DA \times 40\sqrt{3}\) or \(\frac{1}{2} \times 40\sqrt{3} \times 80 \times \sin(\text{their } DCA)\) | M1 | s.o.i. |
| Area of triangle \(= 800\sqrt{3}\) or \(1385.64...\) to 3 s.f. or more | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Area of \(\frac{1}{4}\) circle \(= \frac{1}{2} \times \frac{\pi}{2} \times (40\sqrt{3})^2\) | M1 | \([= 3769.9...]\) |
| \(``6702" + ``1385.6" - ``3769.9"\) | M1 | i.e. their (b)(i) + their (b)(ii) \(-\) their \(\frac{1}{4}\) circle |
| \(= 4300\) to \(4320\) | A1 | \(933\frac{1}{3}\pi + 800\sqrt{3}\) |
## Question 4(a):
$10.6^2 + 9.2^2 - 2 \times 10.6 \times 9.2 \times \cos 68°$ | M1 |
$QR = 11.1(3...)$ | A1 |
$\frac{\sin 68}{\text{their } QR} = \frac{\sin Q}{9.2}$ or $\frac{\sin R}{10.6}$ | M1 | Or correct use of Cosine Rule
$Q = 50.01...°$ or $R = 61.98...°$ | A1 | 2 s.f. or better
Bearing $= 174.9$ to $175°$ | B1 |
---
## Question 4(b)(i):
$(A)\ \frac{1}{2} \times 80^2 \times \frac{2\pi}{3}$ | M1 |
$= \frac{6400\pi}{3}$ | A1 | $6702.(...)$ to 2 s.f. or more
---
## Question 4(b)(ii):
$DC = 80\sin\!\left(\frac{\pi}{3}\right) = 80\frac{\sqrt{3}}{2}$ | B1 | Both steps required
Area $= \frac{1}{2} \times \text{their } DA \times 40\sqrt{3}$ or $\frac{1}{2} \times 40\sqrt{3} \times 80 \times \sin(\text{their } DCA)$ | M1 | s.o.i.
Area of triangle $= 800\sqrt{3}$ or $1385.64...$ to 3 s.f. or more | A1 |
---
## Question 4(b)(iii):
Area of $\frac{1}{4}$ circle $= \frac{1}{2} \times \frac{\pi}{2} \times (40\sqrt{3})^2$ | M1 | $[= 3769.9...]$
$``6702" + ``1385.6" - ``3769.9"$ | M1 | i.e. their (b)(i) + their (b)(ii) $-$ their $\frac{1}{4}$ circle
$= 4300$ to $4320$ | A1 | $933\frac{1}{3}\pi + 800\sqrt{3}$
---4
\begin{enumerate}[label=(\alph*)]
\item \begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e97df57f-3b69-4bec-bc58-9730873dea53-4_765_757_203_764}
\captionsetup{labelformat=empty}
\caption{Fig. 11.1}
\end{center}
\end{figure}
A boat travels from P to Q and then to R . As shown in Fig. 11.1, Q is 10.6 km from P on a bearing of $045 ^ { \circ }$. R is 9.2 km from P on a bearing of $113 ^ { \circ }$, so that angle QPR is $68 ^ { \circ }$.
Calculate the distance and bearing of R from Q .
\item Fig. 11.2 shows the cross-section, EBC, of the rudder of a boat.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e97df57f-3b69-4bec-bc58-9730873dea53-4_527_1474_1452_404}
\captionsetup{labelformat=empty}
\caption{Fig. 11.2}
\end{center}
\end{figure}
BC is an arc of a circle with centre A and radius 80 cm . Angle $\mathrm { CAB } = \frac { 2 \pi } { 3 }$ radians.\\
EC is an arc of a circle with centre D and radius $r \mathrm {~cm}$. Angle CDE is a right angle.
\begin{enumerate}[label=(\roman*)]
\item Calculate the area of sector ABC .
\item Show that $r = 40 \sqrt { 3 }$ and calculate the area of triangle CDA.
\item Hence calculate the area of cross-section of the rudder.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e97df57f-3b69-4bec-bc58-9730873dea53-5_695_1012_271_600}
\captionsetup{labelformat=empty}
\caption{Fig. 12}
\end{center}
\end{figure}
A water trough is a prism 2.5 m long. Fig. 12 shows the cross-section of the trough, with the depths in metres at 0.1 m intervals across the trough. The trough is full of water.
\begin{enumerate}[label=(\roman*)]
\item Use the trapezium rule with 5 strips to calculate an estimate of the area of cross-section of the trough.
Hence estimate the volume of water in the trough.
\item A computer program models the curve of the base of the trough, with axes as shown and units in metres, using the equation $y = 8 x ^ { 3 } - 3 x ^ { 2 } - 0.5 x - 0.15$, for $0 \leqslant x \leqslant 0.5$.
Calculate $\int _ { 0 } ^ { 0.5 } \left( 8 x ^ { 3 } - 3 x ^ { 2 } - 0.5 x - 0.15 \right) \mathrm { d } x$ and state what this represents.\\
Hence find the volume of water in the trough as given by this model.
\end{enumerate}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI C2 Q4 [13]}}