| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| 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 standard bearings problem requiring angle calculation from bearings and application of the sine rule. Part (i) involves straightforward bearing angle conversions, part (ii) is a guided calculation with the answer provided, and part (iii) requires working backwards using given speed. The multi-step nature and bearings context add slight complexity, but the techniques are routine for C2 level with clear scaffolding throughout. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C) |
| Answer | Marks | Guidance |
|---|---|---|
| Angle LAB \(= 60^0\), Angle LBA \(= 70^0\) | B1 B1 | Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Sin rule: \(\frac{LB}{\sin 60} = \frac{5}{\sin 70}\) | M1 A1 | |
| \(\Rightarrow LB = \frac{5 \times \sin 60}{\sin 70} = 4.61\) km (2 d.p.) | A1 | |
| Cosine rule: \(AB^2 = 5^2 + 4.608^2 - 2\times5\times4.608\times\cos 50\) | M1 A1 | |
| \(\Rightarrow AB^2 = 16.614 \Rightarrow AB = 4.076\) km | A1 A1 | |
| \(\Rightarrow\) speed \(= 4.076 \times 2 = 8.15\) km per hour | E1 | Total: 8 |
| Answer | Marks | Guidance |
|---|---|---|
| AB should be 5 km not 4.08 km. That means it should be a parallel line further out. | B1 | |
| \(LA = \frac{5}{4.08} \times 5 = 6.13\) km | B1 | Total: 2 |
## Question 10(i):
Angle LAB $= 60^0$, Angle LBA $= 70^0$ | B1 B1 | **Total: 2**
## Question 10(ii):
Sin rule: $\frac{LB}{\sin 60} = \frac{5}{\sin 70}$ | M1 A1 |
$\Rightarrow LB = \frac{5 \times \sin 60}{\sin 70} = 4.61$ km (2 d.p.) | A1 |
Cosine rule: $AB^2 = 5^2 + 4.608^2 - 2\times5\times4.608\times\cos 50$ | M1 A1 |
$\Rightarrow AB^2 = 16.614 \Rightarrow AB = 4.076$ km | A1 A1 |
$\Rightarrow$ speed $= 4.076 \times 2 = 8.15$ km per hour | E1 | **Total: 8**
## Question 10(iii):
AB should be 5 km not 4.08 km. That means it should be a parallel line further out. | B1 |
$LA = \frac{5}{4.08} \times 5 = 6.13$ km | B1 | **Total: 2**
---10 At 1200 the captain of a ship observes that the bearing of a lighthouse is $340 ^ { \circ }$. His position is at A.\\
At 1230 he takes another bearing of the lighthouse and finds it to be $030 ^ { \circ }$. During this time the ship moves on a constant course of $280 ^ { \circ }$ to the point B .
His plot on the chart is as shown in Fig. 11 below.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{73d1c02b-1b7b-426d-a171-c762597cfed4-3_501_1156_661_387}
\captionsetup{labelformat=empty}
\caption{Fig. 11}
\end{center}
\end{figure}
(i) Write down the size of the angles LAB and LBA .\\
(ii) The captain believes that at A he is 5 km from L . Assuming that LA is exactly 5 km , show that LB is 4.61 km , correct to 2 decimal places, and find AB . Hence calculate the speed of the ship.\\
(iii) The speed of the ship is actually 10 kilometres per hour. Given that the bearings of $340 ^ { \circ }$ and $030 ^ { \circ }$ and the ship's course of $280 ^ { \circ }$ are all accurate, calculate the true value of the distance LA.
\hfill \mbox{\textit{OCR MEI C2 Q10 [12]}}