| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2019 |
| Session | January |
| Marks | 11 |
| 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 testing basic trigonometry (sine/cosine rules), arc length, and sector area formulas. All parts follow standard procedures with no novel insight required, though it involves multiple steps and careful calculation. Slightly easier than average due to the routine nature of each component. |
| 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 |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{\sin DAO}{1.8} = \frac{\sin 0.84}{3.9} \Rightarrow DAO = \text{awrt } 0.351 \text{ rads}\) | M1A1 | M1: Uses sin rule with sides and angles in correct position. A1: Proceeds correctly to \(DAo = \text{awrt } 0.351\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Angle \(ADO = \pi - 0.84 - \text{'0.351'} = (1.95)\) | M1 | States or uses \(ADO = \pi - 0.84 - \text{their } 0.351\) |
| Uses \(AO^2 = 1.8^2 + 3.9^2 - 2\times1.8\times3.9\cos\text{'1.95'}\) | M1 | Uses cosine rule correctly |
| \(AO = 4.9 \text{ (m)}\) | A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Area of sector \(DOBCD = \frac{1}{2}\times1.8^2\times(2\pi - 0.84) = (\text{awrt } 8.8)\) | M1 | |
| Area of triangle \(DOA = \frac{1}{2}\times1.8\times4.9\times\sin 0.84 = (\text{awrt } 3.3)\) | M1 | |
| Area of shop sign \(= \text{awrt } 12.1\ (m^2)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Arc \(BCD = r\theta = 1.8\times(2\pi - 0.84) = (9.8)\) | M1 | |
| Attempts their \(\text{'9.8'} + 3.9 + 3.1\) | M1 | |
| Perimeter of shop sign \(= \text{awrt } 16.8 \text{ (m)}\) | A1 |
## Question 10:
**Part (a):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{\sin DAO}{1.8} = \frac{\sin 0.84}{3.9} \Rightarrow DAO = \text{awrt } 0.351 \text{ rads}$ | M1A1 | M1: Uses sin rule with sides and angles in correct position. A1: Proceeds correctly to $DAo = \text{awrt } 0.351$ |
**Part (b):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| Angle $ADO = \pi - 0.84 - \text{'0.351'} = (1.95)$ | M1 | States or uses $ADO = \pi - 0.84 - \text{their } 0.351$ |
| Uses $AO^2 = 1.8^2 + 3.9^2 - 2\times1.8\times3.9\cos\text{'1.95'}$ | M1 | Uses cosine rule correctly |
| $AO = 4.9 \text{ (m)}$ | A1* | |
**Part (c):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| Area of sector $DOBCD = \frac{1}{2}\times1.8^2\times(2\pi - 0.84) = (\text{awrt } 8.8)$ | M1 | |
| Area of triangle $DOA = \frac{1}{2}\times1.8\times4.9\times\sin 0.84 = (\text{awrt } 3.3)$ | M1 | |
| Area of shop sign $= \text{awrt } 12.1\ (m^2)$ | A1 | |
**Part (d):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| Arc $BCD = r\theta = 1.8\times(2\pi - 0.84) = (9.8)$ | M1 | |
| Attempts their $\text{'9.8'} + 3.9 + 3.1$ | M1 | |
| Perimeter of shop sign $= \text{awrt } 16.8 \text{ (m)}$ | A1 | |10.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{75d68987-2314-4c8f-8160-24977c5c4e34-20_761_1475_331_239}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows the design for a shop sign $A B C D A$.
The sign consists of a triangle $A O D$ joined to a sector of a circle $D O B C D$ with radius 1.8 m and centre $O$.
The points $A , B$ and $O$ lie on a straight line.\\
Given that $A D = 3.9 \mathrm {~m}$ and angle $B O D$ is 0.84 radians,
\begin{enumerate}[label=(\alph*)]
\item calculate the size of angle $D A O$, giving your answer in radians to 3 decimal places.
\item Show that, to one decimal place, the length of $A O$ is 4.9 m .
\item Find, in $\mathrm { m } ^ { 2 }$, the area of the shop sign, giving your answer to one decimal place.
\item Find, in m , the perimeter of the shop sign, giving your answer to one decimal place.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2019 Q10 [11]}}