| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Triangle with circular sector |
| Difficulty | Standard +0.3 This is a straightforward application of cosine rule to find an angle, followed by standard arc length and sector area formulas. All steps are routine with clear signposting - slightly easier than average due to the structured parts and direct application of standard formulas without requiring problem-solving insight. |
| 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 |
| \(\cos\angle BAC = \frac{12^2+10^2-6^2}{2\times12\times10} \Rightarrow \angle BAC = 0.5223\) | M1, A1 | Attempts use of \(6^2=10^2+12^2-2\times10\times12\cos A\); sides must be in correct position; condone different notation e.g. \(\theta\); angle in degrees (awrt \(29.9°\)) is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Arc \(BD = r\theta = 10\times0.5223\) | M1 | Attempts arc formula; in degrees uses Arc \(BD = \frac{\theta}{360}\times2\pi r = \frac{"29.9"}{360}\times2\pi\times10\) |
| Perimeter \(= 6+2+10\times0.5223 = 13.22\) (m) | dM1, A1 | Dependent on arc formula; for calculating perimeter as \(8 +\) arc length; awrt \(13.22\)(m) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Area of sector \(BAD = \frac{1}{2}r^2\theta = \frac{1}{2}\times10^2\times0.5223\ (=26.116)\) | M1 | Attempts area of sector formula; in degrees uses \(\frac{\theta}{360}\times\pi r^2 = \frac{"29.9"}{360}\times\pi\times10^2\) |
| Area of triangle \(ABC = \frac{1}{2}ab\sin C = \frac{1}{2}\times12\times10\times\sin0.5223\ (=29.932)\) | M1 | Attempts area of triangle formula; Heron's formula also accepted with \(S=\frac{10+6+12}{2}=(14)\) and \(A=\sqrt{S(S-10)(S-6)(S-12)}\) |
| Area of flowerbed \(BCD = \frac{1}{2}\times12\times10\times\sin0.5223 - \frac{1}{2}\times10^2\times0.5223\) | dM1 | Dependent on both correct formulae; for finding area of triangle \(-\) area of sector |
| \(= 3.81 / 3.82\ \text{m}^2\) | A1 | Allow awrt \(3.81\) or \(3.82\ \text{m}^2\) |
# Question 6 (Sectors and Triangles):
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\cos\angle BAC = \frac{12^2+10^2-6^2}{2\times12\times10} \Rightarrow \angle BAC = 0.5223$ | M1, A1 | Attempts use of $6^2=10^2+12^2-2\times10\times12\cos A$; sides must be in correct position; condone different notation e.g. $\theta$; angle in degrees (awrt $29.9°$) is A0 | **(2)** |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Arc $BD = r\theta = 10\times0.5223$ | M1 | Attempts arc formula; in degrees uses Arc $BD = \frac{\theta}{360}\times2\pi r = \frac{"29.9"}{360}\times2\pi\times10$ |
| Perimeter $= 6+2+10\times0.5223 = 13.22$ (m) | dM1, A1 | Dependent on arc formula; for calculating perimeter as $8 +$ arc length; awrt $13.22$(m) | **(3)** |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Area of sector $BAD = \frac{1}{2}r^2\theta = \frac{1}{2}\times10^2\times0.5223\ (=26.116)$ | M1 | Attempts area of sector formula; in degrees uses $\frac{\theta}{360}\times\pi r^2 = \frac{"29.9"}{360}\times\pi\times10^2$ |
| Area of triangle $ABC = \frac{1}{2}ab\sin C = \frac{1}{2}\times12\times10\times\sin0.5223\ (=29.932)$ | M1 | Attempts area of triangle formula; Heron's formula also accepted with $S=\frac{10+6+12}{2}=(14)$ and $A=\sqrt{S(S-10)(S-6)(S-12)}$ |
| Area of flowerbed $BCD = \frac{1}{2}\times12\times10\times\sin0.5223 - \frac{1}{2}\times10^2\times0.5223$ | dM1 | Dependent on both correct formulae; for finding area of triangle $-$ area of sector |
| $= 3.81 / 3.82\ \text{m}^2$ | A1 | Allow awrt $3.81$ or $3.82\ \text{m}^2$ | **(4)(9 marks)** |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{08b1be3e-2d9a-4832-b230-d5519540f494-16_364_689_214_630}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of a design for a triangular garden $A B C$.
The garden has sides $B A$ with length $10 \mathrm {~m} , B C$ with length 6 m and $C A$ with length 12 m . The point $D$ lies on $A C$ such that $B D$ is an arc of the circle centre $A$, radius 10 m .
A flowerbed $B C D$ is shown shaded in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Find the size of angle $B A C$, in radians, to 4 decimal places.
\item Find the perimeter of the flowerbed $B C D$, in m , to 2 decimal places.
\item Find the area of the flowerbed $B C D$, in $\mathrm { m } ^ { 2 }$, to 2 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2017 Q6 [9]}}