| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Shaded region with arc |
| Difficulty | Standard +0.3 This is a straightforward C2 question combining cosine rule (part a) with sector/triangle area calculation (part b). Part (a) is routine application of cosine rule with calculator work. Part (b) requires finding shaded area (sector minus triangle) and converting to grass seed amount - standard multi-step procedure with no novel insight required. 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 | Mark | Guidance |
| Way 1: \(10^2 = 7^2 + 13^2 - 2 \times 7 \times 13\cos\theta\) or \(\cos\theta = \frac{7^2 + 13^2 - 10^2}{2 \times 7 \times 13}\) | M1 | Use correct cosine formula in any form |
| \(\cos\theta = \frac{59}{91}\) or \(\cos\theta = 0.6483\) or \(0.8644\) | A1 o.e. | Give a value for \(\cos\theta\) |
| \((\theta = 0.8653789549\ldots) = 0.865^*\) (to 3 dp) | A1* cso | Deduce and state the printed answer \(\theta = 0.865\) |
| Way 2: Uses \(\cos\theta = \frac{x}{7}\), where \(7^2 - x^2 = 10^2 - (13-x)^2\) and finds \(x \ (= 59/13)\) | M1 | |
| \(\cos\theta = \frac{59}{91}\) and \((\theta = 0.8653789549\ldots) = 0.865^*\) (to 3 dp) | A1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area triangle \(ABC = \frac{1}{2} \times 13 \times 7\sin 0.865\) or \(\frac{1}{2} \times 13 \times 7\sin 49.6\) or \(20\sqrt{3}\) | M1 | Correct method for area of correct triangle \(ABC\) |
| Area sector \(ABD = \frac{1}{2} \times 7^2 \times 0.865\) or \(\frac{49.6}{360} \times \pi \times 7^2\) | M1 | Correct method for area of sector |
| \(= 34.6\) (triangle) or \(21.2\) (sector) | A1 | May be implied if not calculated but final answer correct |
| Area of \(S = \frac{1}{2} \times 13 \times 7\sin 0.865 - \frac{1}{2} \times 7^2 \times 0.865 \ (= 13.4)\) | M1 A1 | Correct expression or awrt 13.4 or 13.5 |
| (Amount of seed \(=\)) \(13.4 \times 50 = 670\text{g}\) or \(680\text{g}\) | M1 A1 | Multiply previous answer by 50; 670g or 680g |
# Question 8:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| **Way 1:** $10^2 = 7^2 + 13^2 - 2 \times 7 \times 13\cos\theta$ or $\cos\theta = \frac{7^2 + 13^2 - 10^2}{2 \times 7 \times 13}$ | M1 | Use correct cosine formula in any form |
| $\cos\theta = \frac{59}{91}$ or $\cos\theta = 0.6483$ or $0.8644$ | A1 o.e. | Give a value for $\cos\theta$ |
| $(\theta = 0.8653789549\ldots) = 0.865^*$ (to 3 dp) | A1* cso | Deduce and state the printed answer $\theta = 0.865$ |
| **Way 2:** Uses $\cos\theta = \frac{x}{7}$, where $7^2 - x^2 = 10^2 - (13-x)^2$ and finds $x \ (= 59/13)$ | M1 | |
| $\cos\theta = \frac{59}{91}$ and $(\theta = 0.8653789549\ldots) = 0.865^*$ (to 3 dp) | A1, A1 | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area triangle $ABC = \frac{1}{2} \times 13 \times 7\sin 0.865$ or $\frac{1}{2} \times 13 \times 7\sin 49.6$ or $20\sqrt{3}$ | M1 | Correct method for area of correct triangle $ABC$ |
| Area sector $ABD = \frac{1}{2} \times 7^2 \times 0.865$ or $\frac{49.6}{360} \times \pi \times 7^2$ | M1 | Correct method for area of sector |
| $= 34.6$ (triangle) **or** $21.2$ (sector) | A1 | May be implied if not calculated but final answer correct |
| Area of $S = \frac{1}{2} \times 13 \times 7\sin 0.865 - \frac{1}{2} \times 7^2 \times 0.865 \ (= 13.4)$ | M1 A1 | Correct expression or awrt 13.4 or 13.5 |
| (Amount of seed $=$) $13.4 \times 50 = 670\text{g}$ or $680\text{g}$ | M1 A1 | Multiply previous answer by 50; 670g or 680g |
---8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4f4eac7b-8908-480f-bb39-049944203fff-12_556_1392_210_283}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows the design for a triangular garden $A B C$ where $A B = 7 \mathrm {~m} , A C = 13 \mathrm {~m}$ and $B C = 10 \mathrm {~m}$.
Given that angle $B A C = \theta$ radians,
\begin{enumerate}[label=(\alph*)]
\item show that, to 3 decimal places, $\theta = 0.865$
The point $D$ lies on $A C$ such that $B D$ is an arc of the circle centre $A$, radius 7 m .\\
The shaded region $S$ is bounded by the arc $B D$ and the lines $B C$ and $D C$. The shaded region $S$ will be sown with grass seed, to make a lawned area.
Given that 50 g of grass seed are needed for each square metre of lawn,
\item find the amount of grass seed needed, giving your answer to the nearest 10 g .
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2013 Q8 [10]}}