| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2006 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Segment area calculation |
| Difficulty | Moderate -0.3 This is a standard C2 sector problem with straightforward application of cosine rule, arc length, and area formulas. Part (a) is guided (showing a given result), and all subsequent parts follow directly from standard techniques with no problem-solving insight required. Slightly easier than average due to the scaffolding. |
| 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 theta1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks |
|---|---|
| (a) \(\cos A\hat{O}B = \frac{5^2+5^2-6^2}{2 \times 5 \times 5}\) or | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \frac{7}{25}\) * | A1 cso | (2 marks) |
| (b) \(A\hat{O}B = 1.2870022\ldots\) radians | 1.287 or better | B1 |
| (c) Sector \(= \frac{1}{2} \times 5^2 \times (b)\), \(= 16.087\ldots\) | (AWRT) 16.1 | M1 A1 |
| (d) Triangle \(= \frac{1}{2} \times 5^2 \times \sin(b)\) or \(\frac{1}{2} \times 6 \times \sqrt{5^2-3^2}\) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(=\) (sector from c) \(- 12 = \) (AWRT) 4.1 | dM1; A1 ft (ft their part(c)) | (3 marks) |
(a) $\cos A\hat{O}B = \frac{5^2+5^2-6^2}{2 \times 5 \times 5}$ or | M1
$\sin\theta = \frac{3}{5}$ with use of $\cos 2\theta = 1 - 2\sin^2\theta$ attempted
$= \frac{7}{25}$ * | A1 cso | (2 marks)
(b) $A\hat{O}B = 1.2870022\ldots$ radians | 1.287 or better | B1 | (1 mark)
(c) Sector $= \frac{1}{2} \times 5^2 \times (b)$, $= 16.087\ldots$ | (AWRT) 16.1 | M1 A1 | (2 marks)
(d) Triangle $= \frac{1}{2} \times 5^2 \times \sin(b)$ or $\frac{1}{2} \times 6 \times \sqrt{5^2-3^2}$ | M1
Segment $=$ (their sector) $-$ their triangle
$=$ (sector from c) $- 12 = $ (AWRT) 4.1 | dM1; A1 ft (ft their part(c)) | (3 marks)
**Guidance:**
- (a) M1 for a full method leading to $\cos A\hat{O}B$. [N.B. Use of calculator is M0] (usual rules about quoting formulae)
- (b) Use of (b) in degrees is M0
- (d) 1st M1 for full method for the area of triangle $AOB$; 2nd M1 for their sector $-$ their triangle. Dependent on 1st M1 in part (d). A1 ft for their sector from part (c) $-$ 12 [or 4.1 following a correct restart].
**Total: 8 marks**
---5.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{84b2d36b-c112-4d35-84a1-bc2b707f162d-07_538_611_301_680}
\end{center}
\end{figure}
In Figure $2 O A B$ is a sector of a circle radius 5 m . The chord $A B$ is 6 m long.
\begin{enumerate}[label=(\alph*)]
\item Show that $\cos A \hat { O } B = \frac { 7 } { 25 }$.
\item Hence find the angle $A \hat { O } B$ in radians, giving your answer to 3 decimal places.
\item Calculate the area of the sector $O A B$.
\item Hence calculate the shaded area.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2006 Q5 [8]}}