| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Quadrilateral with diagonal |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard C2 techniques: cosine rule to find a side and angle, then using those results in further calculations, plus a sector area problem. All steps are routine applications of formulas with no novel problem-solving required, making it 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 |
|---|---|---|
| \(AC^2 = 12.8^2 + 7.5^2\) oe; \(AC = 14.83543056..\); \(\tan C = \frac{12.8}{7.5}\); or \(C = 90 - \tan^{-1}(\frac{7.5}{12.8})\) oe; \(59.6\) to \(59.64\); \(\frac{AD}{\sin(155 - \text{their}59.6)} = \frac{\text{their}14.8}{\sin 35}\) oe; \(25.69\) to \(25.8\) | M1, A1, M1, A1, M1, A1 | allow correct application of cosine rule or from finding relevant angle and using trig; rot to 3 or more sf , or 15; B2 for 14.8 or better unsupported; or \(\sin C = \frac{12.8}{\text{their}14.8}\); or \(\cos C = \frac{7.5}{\text{their}14.8}\); or \(\cos C = \frac{\text{their}14.8^2 + 7.5^2 - 12.8^2}{2 \times 7.5 \times \text{their}14.8}\); allow B2 for 25.69 \(\leq AD < 25.8\) unsupported....but B0 for 25.8 unsupported; M0A0 for \(\frac{14.9}{\text{cos}35} = 25.803...\) |
| Answer | Marks | Guidance |
|---|---|---|
| area of \(ABC = 48\) soi; \(\frac{1}{2} \times \text{their} 14.8 \times ...\text{their} 25.7...\times \sin(\text{their} 59.6 - 10)\); \(192.8\) to \(194[\text{m}^2]\) | B1, M1, A1 | may be implied by correct final answer in range or by sight of \(\frac{1}{2} \times 12.8 \times 7.5\) oe; may be implied by 144.8 to 146; B3 for correct answer in range if unsupported; condone 48.0... |
| Answer | Marks | Guidance |
|---|---|---|
| angle \(HMG = \frac{\pi - 1.1}{2}\); or \(MHG = 0.55\) (31.5126°); \(HM = 1.7176\) to \(1.7225\); \(\frac{1}{2} \times 1.1 \times \text{their } HM^2\); or \(\frac{\theta}{360} \times \pi \times \text{their}HM^2\); area or triangle \(EMF = 0.652\) to \(0.662\); \(2.95\) to \(2.952\) [m²] cao | B1, B1, M1, B1, A1 | or angle \(EMF\); or angle \(MEF\); may be implied by final answer; 1.63(0661924...); \(\theta=63(.025357...)\); or \(MGH\); may be implied by final answer or in double this (1.304 to 1.324); full marks may be awarded for final answer in correct range ie allow recovery of accuracy; check arithmetic if necessary their \(HM \neq 0.9\) or 1.8 |
| Answer | Marks | Guidance |
|---|---|---|
| \(65 \times (1 - 0.017)^3\) oe | M1 | may be longer method finding decrease year by year etc |
| \(61.7410...\) showing more than 3 sf | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \([d =] 65 \times 0.983°\) oe | B1 | eg 63.895 × 0.983^{-1} or 61.7 × 0.983^{-3} |
### Part (i) (A)
$AC^2 = 12.8^2 + 7.5^2$ oe; $AC = 14.83543056..$; $\tan C = \frac{12.8}{7.5}$; or $C = 90 - \tan^{-1}(\frac{7.5}{12.8})$ oe; $59.6$ to $59.64$; $\frac{AD}{\sin(155 - \text{their}59.6)} = \frac{\text{their}14.8}{\sin 35}$ oe; $25.69$ to $25.8$ | M1, A1, M1, A1, M1, A1 | allow correct application of cosine rule or from finding relevant angle and using trig; rot to 3 or more sf , or 15; B2 for 14.8 or better unsupported; or $\sin C = \frac{12.8}{\text{their}14.8}$; or $\cos C = \frac{7.5}{\text{their}14.8}$; or $\cos C = \frac{\text{their}14.8^2 + 7.5^2 - 12.8^2}{2 \times 7.5 \times \text{their}14.8}$; allow B2 for 25.69 $\leq AD < 25.8$ unsupported....but B0 for 25.8 unsupported; M0A0 for $\frac{14.9}{\text{cos}35} = 25.803...$
### Part (i) (B)
area of $ABC = 48$ soi; $\frac{1}{2} \times \text{their} 14.8 \times ...\text{their} 25.7...\times \sin(\text{their} 59.6 - 10)$; $192.8$ to $194[\text{m}^2]$ | B1, M1, A1 | may be implied by correct final answer in range or by sight of $\frac{1}{2} \times 12.8 \times 7.5$ oe; may be implied by 144.8 to 146; B3 for correct answer in range if unsupported; condone 48.0...
### Part (ii)
angle $HMG = \frac{\pi - 1.1}{2}$; or $MHG = 0.55$ (31.5126°); $HM = 1.7176$ to $1.7225$; $\frac{1}{2} \times 1.1 \times \text{their } HM^2$; or $\frac{\theta}{360} \times \pi \times \text{their}HM^2$; area or triangle $EMF = 0.652$ to $0.662$; $2.95$ to $2.952$ [m²] cao | B1, B1, M1, B1, A1 | or angle $EMF$; or angle $MEF$; may be implied by final answer; 1.63(0661924...); $\theta=63(.025357...)$; or $MGH$; may be implied by final answer or in double this (1.304 to 1.324); full marks may be awarded for final answer in correct range ie allow recovery of accuracy; check arithmetic if necessary their $HM \neq 0.9$ or 1.8
### Part (i)
$65 \times (1 - 0.017)^3$ oe | M1 | may be longer method finding decrease year by year etc | NB use of $3 \times 0.017$ leads to 61.685, which doesn't score
$61.7410...$ showing more than 3 sf | A1
### Part (ii)
$[d =] 65 \times 0.983°$ oe | B1 | eg 63.895 × 0.983^{-1} or 61.7 × 0.983^{-3}Fig. 10.1 shows Jean's back garden. This is a quadrilateral ABCD with dimensions as shown.
\includegraphics{figure_10.1}
\begin{enumerate}[label=(\roman*)]
\item \begin{enumerate}[label=(\alph*)]
\item Calculate AC and angle ACB. Hence calculate AD. [6]
\item Calculate the area of the garden. [3]
\end{enumerate}
\item The shape of the fence panels used in the garden is shown in Fig. 10.2. EH is the arc of a sector of a circle with centre at the midpoint, M, of side FG, and sector angle 1.1 radians, as shown. FG = 1.8 m.
\includegraphics{figure_10.2}
Calculate the area of one of these fence panels. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C2 2013 Q10 [14]}}