| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2010 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Sector with attached triangle |
| Difficulty | Standard +0.3 This is a straightforward multi-part question combining basic triangle trigonometry (sine rule) with standard sector area formula. Part (a) guides students through finding an angle, and part (b) requires adding triangle and sector areas using bookwork formulas. Slightly easier than average due to clear structure and standard techniques. |
| 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Either \(\dfrac{\sin(A\hat{C}B)}{5} = \dfrac{\sin 0.6}{4}\) | M1 | Correct use of sine rule to find \(ACB\); (M0 for use of \(\sin x\) where \(x\) could be \(ABC\)) |
| \(\therefore A\hat{C}B = \arcsin(0.7058\ldots) = [0.7835\ldots \text{ or } 2.358]\) | M1 | Correct expression for angle \(ACB\) |
| Use angles of triangle; \(A\hat{B}C = \pi - 0.6 - A\hat{C}B\) | M1 | Correct method to get angle \(ABC\) |
| \(A\hat{B}C = 1.76\) (3sf) [Allow \(100.7° \to 1.76\)] | A1 | Correct work leading to 1.76 (3sf) |
| Or \(4^2 = b^2 + 5^2 - 2 \times b \times 5\cos 0.6\) | M1 | Correct use of cosine rule to find \(b\) |
| \(b = \dfrac{10\cos 0.6 \pm \sqrt{100\cos^2 0.6 - 36}}{2} = [6.96 \text{ or } 1.29]\) | M1 | Correct expression for \(b\) |
| \(\sin B = \dfrac{\sin 0.6}{4} \times b\) or \(\cos B = \dfrac{25+16-b^2}{40}\); \(A\hat{B}C = 1.76\) (3sf) | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(C\hat{B}D = \pi - 1.76 = 1.38\); Sector area \(= \frac{1}{2} \times 4^2 \times (\pi - 1.76) = [11.0\text{ to }11.1]\) | M1 | Correct expression for sector area or value in range \(11.0\)–\(11.1\) |
| Area of \(\triangle ABC = \frac{1}{2} \times 5 \times 4 \times \sin(1.76) = [9.8]\) | M1 | Correct expression for area of triangle or value of 9.8 |
| Required area \(\approx 20.8\) or \(20.9\) or \(21.0\) | A1 | Answers which round to 20.8, 20.9, or 21.0 |
## Question 4:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Either $\dfrac{\sin(A\hat{C}B)}{5} = \dfrac{\sin 0.6}{4}$ | M1 | Correct use of sine rule to find $ACB$; (M0 for use of $\sin x$ where $x$ could be $ABC$) |
| $\therefore A\hat{C}B = \arcsin(0.7058\ldots) = [0.7835\ldots \text{ or } 2.358]$ | M1 | Correct expression for angle $ACB$ |
| Use angles of triangle; $A\hat{B}C = \pi - 0.6 - A\hat{C}B$ | M1 | Correct method to get angle $ABC$ |
| $A\hat{B}C = 1.76$ (3sf) [Allow $100.7° \to 1.76$] | A1 | Correct work leading to 1.76 (3sf) |
| Or $4^2 = b^2 + 5^2 - 2 \times b \times 5\cos 0.6$ | M1 | Correct use of cosine rule to find $b$ |
| $b = \dfrac{10\cos 0.6 \pm \sqrt{100\cos^2 0.6 - 36}}{2} = [6.96 \text{ or } 1.29]$ | M1 | Correct expression for $b$ |
| $\sin B = \dfrac{\sin 0.6}{4} \times b$ or $\cos B = \dfrac{25+16-b^2}{40}$; $A\hat{B}C = 1.76$ (3sf) | M1, A1 | |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $C\hat{B}D = \pi - 1.76 = 1.38$; Sector area $= \frac{1}{2} \times 4^2 \times (\pi - 1.76) = [11.0\text{ to }11.1]$ | M1 | Correct expression for sector area or value in range $11.0$–$11.1$ |
| Area of $\triangle ABC = \frac{1}{2} \times 5 \times 4 \times \sin(1.76) = [9.8]$ | M1 | Correct expression for area of triangle or value of 9.8 |
| Required area $\approx 20.8$ or $20.9$ or $21.0$ | A1 | Answers which round to 20.8, 20.9, or 21.0 |4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e3faf018-37a8-48ef-b100-81402a8ec87f-05_556_1189_237_413}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
An emblem, as shown in Figure 1, consists of a triangle $A B C$ joined to a sector $C B D$ of a circle with radius 4 cm and centre $B$. The points $A , B$ and $D$ lie on a straight line with $A B = 5 \mathrm {~cm}$ and $B D = 4 \mathrm {~cm}$. Angle $B A C = 0.6$ radians and $A C$ is the longest side of the triangle $A B C$.
\begin{enumerate}[label=(\alph*)]
\item Show that angle $A B C = 1.76$ radians, correct to 3 significant figures.
\item Find the area of the emblem.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2010 Q4 [7]}}