| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2006 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Bearings and navigation |
| Difficulty | Moderate -0.3 This is a straightforward application of the cosine rule, sine rule, and arc length formula with no conceptual challenges. Part (i) uses standard triangle solving techniques with bearings, while part (ii) involves routine arc length calculation (s=rθ) and bearing adjustment. All steps are procedural with clear methods, making it slightly easier than average for A-level. |
| 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 |
|---|---|---|
| i (A) \(5.2^2 + 6.3^2 - 2 \times 5.2 \times 6.3 \times \cos \text{"57"}\) | M2 | M1 for recognisable attempt at cos rule or greater accuracy |
| \(ST = 5.6\) or \(5.57\) cao | A1 | |
| i (B) \(\sin T/5.2 = \sin(\text{their 57})/\text{their ST}\) | M1 | Or \(\sin S/6.3 = \ldots\), or cosine rule |
| \(T = 51\) to \(52\) or \(S = 71\) to \(72\) | A1 | |
| bearing \(285 +\) their \(T\) or \(408 -\) their \(S\) | B1 | If outside 0 to 360, must be adjusted |
| ii \(5.2\theta, 24 \times 26/60\) | B1B1 | |
| \(\theta = 1.98\) to \(2.02\) | B1 | |
| \(\text{arc} = \text{their 2} \times 180/\pi\) or \(114.6°...\) | M1 | |
| Bearing \(= 293\) to \(294\) cao | A1 | Lost for all working in degrees; Implied by 57.3 |
i (A) $5.2^2 + 6.3^2 - 2 \times 5.2 \times 6.3 \times \cos \text{"57"}$ | M2 | M1 for recognisable attempt at cos rule or greater accuracy |
$ST = 5.6$ or $5.57$ cao | A1 | | $3$ |
i (B) $\sin T/5.2 = \sin(\text{their 57})/\text{their ST}$ | M1 | Or $\sin S/6.3 = \ldots$, or cosine rule |
$T = 51$ to $52$ or $S = 71$ to $72$ | A1 | |
bearing $285 +$ their $T$ or $408 -$ their $S$ | B1 | If outside 0 to 360, must be adjusted | $3$ |
ii $5.2\theta, 24 \times 26/60$ | B1B1 | |
$\theta = 1.98$ to $2.02$ | B1 | |
$\text{arc} = \text{their 2} \times 180/\pi$ or $114.6°...$ | M1 | |
Bearing $= 293$ to $294$ cao | A1 | Lost for all working in degrees; Implied by 57.3 | $5$ | $11$ |\begin{enumerate}[label=(\roman*)]
\item
\includegraphics{figure_10_1}
At a certain time, ship S is 5.2 km from lighthouse L on a bearing of 048°. At the same time, ship T is 6.3 km from L on a bearing of 105°, as shown in Fig. 10.1.
For these positions, calculate
\begin{enumerate}[label=(\Alph*)]
\item the distance between ships S and T, [3]
\item the bearing of S from T. [3]
\end{enumerate}
\item
\includegraphics{figure_10_2}
Ship S then travels at 24 km 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. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C2 2006 Q10 [11]}}