| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2009 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Multiple circles or sectors |
| Difficulty | Standard +0.3 This is a straightforward application of standard arc length and sector area formulas in part (i), followed by basic trigonometry (cosine rule) in part (ii). All steps are routine with clear given information and no conceptual challenges beyond direct formula application, 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(150\) (cm) or \(1.5\) m | 2 | M1 for \(2.5 \times 60\) or \(2.5 \times 0.6\) or for \(1.5\) with no units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2} \times 60^2 \times 2.5\) or \(4500\) | M1 | or equivalents in m² |
| \(\frac{1}{2} \times 140^2 \times 2.5\) or \(24500\) | M1 | |
| subtraction of these | DM1 | |
| \(20\,000\) (cm²) isw | A1 | or \(2\) m² |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| attempt at use of cosine rule | M1 | condone 1 error in substitution |
| \(\cos\text{EFP} = \frac{3.5^2 + 2.8^2 - 1.6^2}{2 \times 2.8 \times 3.5}\) o.e. | M1 | |
| \(26.5\) to \(26.65\) or \(27\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2.8 \sin(\text{their EFP})\) o.e. | M1 | |
| \(1.2\) to \(1.3\) [m] | A1 |
## Question 11:
### Part i(A):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $150$ (cm) or $1.5$ m | 2 | M1 for $2.5 \times 60$ or $2.5 \times 0.6$ or for $1.5$ with no units | **[2]** |
### Part i(B):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2} \times 60^2 \times 2.5$ or $4500$ | M1 | or equivalents in m² |
| $\frac{1}{2} \times 140^2 \times 2.5$ or $24500$ | M1 | |
| subtraction of these | DM1 | |
| $20\,000$ (cm²) isw | A1 | or $2$ m² | **[4]** |
### Part ii(A):
| Answer/Working | Marks | Guidance |
|---|---|---|
| attempt at use of cosine rule | M1 | condone 1 error in substitution |
| $\cos\text{EFP} = \frac{3.5^2 + 2.8^2 - 1.6^2}{2 \times 2.8 \times 3.5}$ o.e. | M1 | |
| $26.5$ to $26.65$ or $27$ | A1 | | **[3]** |
### Part ii(B):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2.8 \sin(\text{their EFP})$ o.e. | M1 | |
| $1.2$ to $1.3$ [m] | A1 | | **[2]** |
---11
\begin{enumerate}[label=(\roman*)]
\item \begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{838d6b15-69a9-4e67-bc36-5bf60254a767-5_469_878_274_671}
\captionsetup{labelformat=empty}
\caption{Fig. 11.1}
\end{center}
\end{figure}
Fig. 11.1 shows the surface ABCD of a TV presenter's desk. AB and CD are arcs of circles with centre O and sector angle 2.5 radians. $\mathrm { OC } = 60 \mathrm {~cm}$ and $\mathrm { OB } = 140 \mathrm {~cm}$.\\
(A) Calculate the length of the arc CD.\\
(B) Calculate the area of the surface ABCD of the desk.
\item The TV presenter is at point P , shown in Fig. 11.2. A TV camera can move along the track EF , which is of length 3.5 m .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{838d6b15-69a9-4e67-bc36-5bf60254a767-5_378_877_1334_675}
\captionsetup{labelformat=empty}
\caption{Fig. 11.2}
\end{center}
\end{figure}
When the camera is at E , the TV presenter is 1.6 m away. When the camera is at F , the TV presenter is 2.8 m away.\\
(A) Calculate, in degrees, the size of angle EFP.\\
(B) Calculate the shortest possible distance between the camera and the TV presenter.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C2 2009 Q11 [11]}}