| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Shaded region with arc |
| Difficulty | Standard +0.3 This is a multi-part question requiring cosine rule (part a), triangle area formula (part b), and sector area with algebraic manipulation (part c). While it involves several steps, each technique is standard C2 material with straightforward application. The final part requires setting up an equation (sector area = remaining area) but the algebra is routine, making this slightly easier than average. |
| 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 |
|---|---|---|
| (a) | \(5^2 = 8^2 + 9^2 - 2(8)(9)\cos \theta\) | M1 |
| (a) | \(\cos \theta = \frac{8^2 + 9^2 - 5^2}{2(8)(9)} \left(= \frac{120}{144} = \frac{5}{6}\right)\) | m1 |
| (a) | \(\theta = 0.5856(855...) = 0.586\) to 3sf | A1 |
| (b) | (Area of triangle) = \(\frac{1}{2} \times 9 \times 8 \sin \theta\) | M1 |
| (b) | \(= 36\sin \theta = 19.9\) (cm²) | A1 |
| (c) | (Area of sector) = \(\frac{1}{2}r^2\theta\) | M1 |
| (c) | \(\frac{1}{2}r^2\theta = 36\sin \theta - \frac{1}{2}r^2\theta\) | m1 |
| (c) | \(r^2 = \frac{36 \sin \theta}{\theta} \approx 34, \quad r = 5.83\) | A1 |
| (c) | (Arc length)= \(r\theta\) | M1 |
| (c) | Perimeter (of shaded shape) \(= r\theta + 5 + 8 - r + 9 - r\) | m1 |
| (c) | \(= 22 + r(\theta - 2)\) | |
| (c) | \(= 13.8\) (cm) to 3sf | A1 |
| Total | 11 |
(a) | $5^2 = 8^2 + 9^2 - 2(8)(9)\cos \theta$ | M1 | Cosine rule used correctly. Accept eg A for $\theta$ if intention is clear. PI by next line. Allow one sign slip in rearrangement from a correct M1 line
(a) | $\cos \theta = \frac{8^2 + 9^2 - 5^2}{2(8)(9)} \left(= \frac{120}{144} = \frac{5}{6}\right)$ | m1 |
(a) | $\theta = 0.5856(855...) = 0.586$ to 3sf | A1 | 3 marks | AG Must see more than 3sf for the angle before seeing printed value 0.586
(b) | (Area of triangle) = $\frac{1}{2} \times 9 \times 8 \sin \theta$ | M1 | OE Correct value or correct expression involving no unknown values. eg $\sqrt{11(11-9)(11-8)(11-5)}$
(b) | $= 36\sin \theta = 19.9$ (cm²) | A1 | 2 marks | If >3sf accept a value from 19.89 to 19.91 inclusive. NMS: Award 2 marks for 19.9 or 'better'
(c) | (Area of sector) = $\frac{1}{2}r^2\theta$ | M1 | $\frac{1}{2}r^2\theta$ seen or used, for sector area
(c) | $\frac{1}{2}r^2\theta = 36\sin \theta - \frac{1}{2}r^2\theta$ | m1 | OE Ft c's answer to (b) if incorrect eg $\frac{1}{2}r^2\theta = $ "19.9"
(c) | $r^2 = \frac{36 \sin \theta}{\theta} \approx 34, \quad r = 5.83$ | A1 | PI by later work. Accept 5.82 to 5.83 inclusive.
(c) | (Arc length)= $r\theta$ | M1 | $r\theta$ seen or used, for arc length
(c) | Perimeter (of shaded shape) $= r\theta + 5 + 8 - r + 9 - r$ | m1 | Correct expression, seen or used, for the required perimeter.
(c) | $= 22 + r(\theta - 2)$ | | CAO must be 13.8
(c) | $= 13.8$ (cm) to 3sf | A1 | 6 marks |
| **Total** | **11** |
**(a)** Accept 0.5856 or 0.5857 or AWFW 0.5856 to 0.5857 as evidence for the A1
**(b)** NMS: 'better' means 'value from 19.89 to 19.91 inclusive'
---6 The diagram shows a triangle $A B C$.
The lengths of $A B , B C$ and $A C$ are $8 \mathrm {~cm} , 5 \mathrm {~cm}$ and 9 cm respectively.\\
Angle $B A C$ is $\theta$ radians.
\begin{enumerate}[label=(\alph*)]
\item Show that $\theta = 0.586$, correct to three significant figures.
\item Find the area of triangle $A B C$, giving your answer, in $\mathrm { cm } ^ { 2 }$, to three significant figures.
\item A circular sector, centre $A$ and radius $r \mathrm {~cm}$, is removed from triangle $A B C$. The remaining shape is shown shaded in the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{e183578a-29a8-4112-b941-06c8894ed078-14_467_677_1462_685}
Given that the area of the sector removed is equal to the area of the shaded shape, find the perimeter of the shaded shape. Give your answer in cm to three significant figures.\\[0pt]
[6 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2016 Q6 [11]}}