| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Finding angle from geometry |
| Difficulty | Standard +0.3 Parts (a) and (b) are direct formula applications (arc length = rθ, area = ½r²θ) requiring simple substitution. Part (c) requires using the sine rule or cosine rule in a triangle, which is slightly more challenging but still a standard C2 technique with straightforward setup given the numerical values. |
| 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 |
|---|---|---|
| \(\{Arc\} = r\theta = 20 \times 0.8 = 16\) (cm) | M1, A1 | \(r\theta\) seen in (a) or used for the arc length |
| Answer | Marks | Guidance |
|---|---|---|
| \(\{Area\ of\ sector\} = \frac{1}{2}r^2\theta = \frac{1}{2} \times 20^2 \times 0.8\) | M1 | \(\frac{1}{2}r^2\theta\) OE seen in (b) or used for the area |
| \(= 160\) (cm²) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\{Let\ D = angle\ ODB\}\) \(\frac{20}{\sin D} = \frac{15}{\sin 0.8}\) | M1 | Sine rule, ACF with sin \(D\) being the only unknown PI by next line |
| \(\sin D = \frac{20 \times \sin 0.8}{15} = \frac{14.3(471...)}{15}\) | m1 | Correct rearrangement to 'sinD = ...' or to 'D=sin\(^{-1}\)(...)' OE. PI by at least 3sf correct value 1.27(467...) radians or 73(.033)° for acute angle \(D\) by at least 3sf value 1.86(692...) rounded or truncated for \(D\). |
| \(\left\{\frac{20}{20.9(10...)}\right\} = 0.956(474...)\) | ||
| \(Acute\ 'D' = 1.27(467...)\) | ||
| \(D = \pi - Acute\ 'D'\) in rads | m1 | Dep on previous 2 marks being awarded. PI by correct ft evaluation of \(\pi - c'\)s acute \(D\) to at least 3 sf value or seeing 1.86(692...), rounded or truncated, for \(D\). |
| \(\{Angle\ ODB\} = 1.87\) [to 3sf] | A1 | Condone >3sf |
| Total | 8 |
**2(a)**
$\{Arc\} = r\theta = 20 \times 0.8 = 16$ (cm) | M1, A1 | $r\theta$ seen in (a) or used for the arc length | 2
**2(b)**
$\{Area\ of\ sector\} = \frac{1}{2}r^2\theta = \frac{1}{2} \times 20^2 \times 0.8$ | M1 | $\frac{1}{2}r^2\theta$ OE seen in (b) or used for the area
$= 160$ (cm²) | A1 | | 2
**2(c)**
$\{Let\ D = angle\ ODB\}$ $\frac{20}{\sin D} = \frac{15}{\sin 0.8}$ | M1 | Sine rule, ACF with sin $D$ being the only unknown PI by next line
$\sin D = \frac{20 \times \sin 0.8}{15} = \frac{14.3(471...)}{15}$ | m1 | Correct rearrangement to 'sinD = ...' or to 'D=sin$^{-1}$(...)' OE. PI by at least 3sf correct value 1.27(467...) radians or 73(.033)° for acute angle $D$ by at least 3sf value 1.86(692...) rounded or truncated for $D$.
$\left\{\frac{20}{20.9(10...)}\right\} = 0.956(474...)$ | |
$Acute\ 'D' = 1.27(467...)$ | |
$D = \pi - Acute\ 'D'$ in rads | m1 | Dep on previous 2 marks being awarded. PI by correct ft evaluation of $\pi - c'$s acute $D$ to at least 3 sf value or seeing 1.86(692...), rounded or truncated, for $D$.
$\{Angle\ ODB\} = 1.87$ [to 3sf] | A1 | Condone >3sf | 4
Total | | 8
---2 The diagram shows a sector $O A B$ of a circle with centre $O$.\\
\includegraphics[max width=\textwidth, alt={}, center]{f4f090a1-7e36-4993-a49e-b6e7e8589057-2_341_371_968_815}
The radius of the circle is 20 cm and the angle $A O B = 0.8$ radians.
\begin{enumerate}[label=(\alph*)]
\item Find the length of the arc $A B$.
\item Find the area of the sector $O A B$.
\item A line from $B$ meets the radius $O A$ at the point $D$, as shown in the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{f4f090a1-7e36-4993-a49e-b6e7e8589057-2_344_371_1747_815}
The length of $B D$ is 15 cm . Find the size of the obtuse angle $O D B$, in radians, giving your answer to three significant figures.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2013 Q2 [8]}}