| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Segment area calculation |
| Difficulty | Moderate -0.8 This is a straightforward C2 question testing standard radian formulas (arc length, sector area, segment area). Part (a) uses cosine rule, parts (b)-(c) apply memorized formulas with minimal problem-solving. The multi-part structure adds length but not conceptual difficulty—each step is routine application of basic techniques. |
| 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 |
|---|---|---|
| \(32^2 = 24^2 + 24^2 - 2 \times 24 \times 24 \cos\theta\); \(\sin\frac{1}{2}\theta = \frac{2}{32}(32)\); \(\cos\theta = \frac{24^2 + 24^2 - 32^2}{2 \times 24 \times 24}\); or \(\frac{1}{2}\theta = \sin^{-1}\left(\frac{2}{3}\right) (= 0.7297...)\); \(\theta = 1.459... = 1.46\) to 3sf | M1, m1, A1 | CSO AG (be convinced) |
| Total for 2(a): 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Arc: \(r\theta = 24 \times 1.459... = 35\) cm | M1, A1 | Condone absent cm; 35 to 35.04 |
| Total for 2(b): 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Area of sector: \(\frac{1}{2}r^2\theta = \frac{1}{2}(24)^2(1.459...) = 420.3 = 420\) cm² | M1, A1 | Seen; Condone absent cm²; 420 to 420.48 |
| Total for 2(c)(i): 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Area of triangle: \(\frac{1}{2}(24)(24)\sin\theta = [= 286.(...)]\); Shaded area = area of sector − area of triangle; \(\left[\frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta\right] = 134\) cm² | M1, m1, A1 | OE; Dep on at least one of the previous two M marks. PI; Condone absent cm² |
| Total for 2(c)(ii): 3 marks | ||
| Total for Question 2: 10 marks |
**2(a)**
| $32^2 = 24^2 + 24^2 - 2 \times 24 \times 24 \cos\theta$; $\sin\frac{1}{2}\theta = \frac{2}{32}(32)$; $\cos\theta = \frac{24^2 + 24^2 - 32^2}{2 \times 24 \times 24}$; or $\frac{1}{2}\theta = \sin^{-1}\left(\frac{2}{3}\right) (= 0.7297...)$; $\theta = 1.459... = 1.46$ to 3sf | M1, m1, A1 | CSO AG (be convinced) |
| Total for 2(a): 3 marks |
**2(b)**
| Arc: $r\theta = 24 \times 1.459... = 35$ cm | M1, A1 | Condone absent cm; 35 to 35.04 |
| Total for 2(b): 2 marks |
**2(c)(i)**
| Area of sector: $\frac{1}{2}r^2\theta = \frac{1}{2}(24)^2(1.459...) = 420.3 = 420$ cm² | M1, A1 | Seen; Condone absent cm²; 420 to 420.48 |
| Total for 2(c)(i): 2 marks |
**2(c)(ii)**
| Area of triangle: $\frac{1}{2}(24)(24)\sin\theta = [= 286.(...)]$; Shaded area = area of sector − area of triangle; $\left[\frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta\right] = 134$ cm² | M1, m1, A1 | OE; Dep on at least one of the previous two M marks. PI; Condone absent cm² |
| Total for 2(c)(ii): 3 marks |
| **Total for Question 2: 10 marks** |
---2 The diagram shows a triangle $A B C$ and the arc $A B$ of a circle whose centre is $C$ and whose radius is 24 cm .\\
\includegraphics[max width=\textwidth, alt={}, center]{4a4d4dcd-4137-427d-834f-ac2fe83f8aeb-2_506_403_1187_781}
The length of the side $A B$ of the triangle is 32 cm . The size of the angle $A C B$ is $\theta$ radians.
\begin{enumerate}[label=(\alph*)]
\item Show that $\theta = 1.46$ correct to three significant figures.
\item Calculate the length of the $\operatorname { arc } A B$ to the nearest cm .
\item \begin{enumerate}[label=(\roman*)]
\item Calculate the area of the sector $A B C$ to the nearest $\mathrm { cm } ^ { 2 }$.
\item Hence calculate the area of the shaded segment to the nearest $\mathrm { cm } ^ { 2 }$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2005 Q2 [10]}}