| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Bearings and navigation |
| Difficulty | Standard +0.3 This is a standard C2 question testing routine applications of sine/cosine rules, bearings, sector area, and basic trigonometry with radians. Part (a) is a straightforward two-step bearing problem (5 marks). Part (b) involves standard sector area formula, a guided calculation showing r = 40√3 using basic trigonometry, and combining areas—all textbook exercises with no novel insight required. Slightly above average difficulty due to the multi-part nature and mark allocation, but well within typical C2 scope. |
| 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 |
|---|---|---|
| \(10.6^2 + 9.2^2 - 2 \times 10.6 \times 9.2\cos 68°\) o.e. | M1 | |
| \(QR = 11.1(3...)\) | A1 | |
| \(\frac{\sin 68}{\text{their }QR} = \frac{\sin Q}{9.2}\) or \(\frac{\sin R}{10.6}\) o.e. | 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 |
|---|---|---|
| \(DC = 80 \sin\left(\frac{2\pi}{3}\right) = 80\frac{\sqrt{3}}{2}\) | B1 | Both steps required |
| Area \(= \frac{1}{2} \times \text{their }DA \times 40\sqrt{3}\) or \(\frac{1}{2} \times 40\sqrt{3} \times 80\sin(\text{their }DCA)\) o.e. | 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 \(= \frac{1}{2} \times \frac{\pi}{2} \times (40\sqrt{3})^2\) o.e. | M1 | \([=3769.9...]\) |
| "\(6702\)" + "\(1385.6\)" − "\(3769.9\)" | M1 | i.e. their(b)(i) + their(b)(ii) − their \(\frac{1}{4}\) circle o.e. |
| \(= 4300\) to \(4320\) | A1 | \(933\frac{1}{3}\pi + 800\sqrt{3}\) |
## (a)
$10.6^2 + 9.2^2 - 2 \times 10.6 \times 9.2\cos 68°$ o.e. | M1 |
$QR = 11.1(3...)$ | A1 |
$\frac{\sin 68}{\text{their }QR} = \frac{\sin Q}{9.2}$ or $\frac{\sin R}{10.6}$ o.e. | 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 |
## (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
## (b)(ii)
$DC = 80 \sin\left(\frac{2\pi}{3}\right) = 80\frac{\sqrt{3}}{2}$ | B1 | Both steps required
Area $= \frac{1}{2} \times \text{their }DA \times 40\sqrt{3}$ or $\frac{1}{2} \times 40\sqrt{3} \times 80\sin(\text{their }DCA)$ o.e. | M1 | s.o.i.
Area of triangle $= 800\sqrt{3}$ or $1385.64...$ to 3 s.f. or more | A1 |
## (b)(iii)
Area of $\frac{1}{4}$ circle $= \frac{1}{2} \times \frac{\pi}{2} \times (40\sqrt{3})^2$ o.e. | M1 | $[=3769.9...]$
"$6702$" + "$1385.6$" − "$3769.9$" | M1 | i.e. their(b)(i) + their(b)(ii) − their $\frac{1}{4}$ circle o.e.
$= 4300$ to $4320$ | A1 | $933\frac{1}{3}\pi + 800\sqrt{3}$\begin{enumerate}[label=(\alph*)]
\item \includegraphics{figure_11_1}
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 $045°$. R is 9.2 km from P on a bearing of $113°$, so that angle QPR is $68°$.
Calculate the distance and bearing of R from Q. [5]
\item Fig. 11.2 shows the cross-section, EBC, of the rudder of a boat.
\includegraphics{figure_11_2}
BC is an arc of a circle with centre A and radius 80 cm. Angle CAB = $\frac{2\pi}{3}$ radians.
EC is an arc of a circle with centre D and radius $r$ cm. Angle CDE is a right angle.
\begin{enumerate}[label=(\roman*)]
\item Calculate the area of sector ABC. [2]
\item Show that $r = 40\sqrt{3}$ and calculate the area of triangle CDA. [3]
\item Hence calculate the area of cross-section of the rudder. [3]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C2 2010 Q11 [13]}}