| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| 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 requiring standard applications of cosine rule, sine rule, and area formulas (triangles and sectors). While it involves multiple steps across two contexts, each individual calculation is routine with clear signposting, making it slightly easier than average for A-level. |
| 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 | Mark | Guidance |
| \(AC^2 = 12.8^2 + 7.5^2\) oe | M1 | Allow correct application of cosine rule or from finding relevant angle and using trig |
| \(AC = 14.83543056...\) | A1 | Rot to 3 or more sf, or 15. B2 for 14.8 or better unsupported |
| \(\tan C = \frac{12.8}{7.5}\) or \(C = 90 - \tan^{-1}\left(\frac{7.5}{12.8}\right)\) oe | M1 | or \(\sin C = \frac{12.8}{\text{their}14.8}\) or \(\cos C = \frac{7.5}{\text{their}14.8}\); or \(\frac{\sin C}{12.8} = \frac{\sin 90}{\text{their}14.8}\); or \(\cos C = \frac{\text{their}14.8^2 + 7.5^2 - 12.8^2}{2 \times 7.5 \times \text{their}14.8}\) |
| 59.6 to 59.64 | A1 | |
| \(\frac{AD}{\sin(155 - \text{their}59.6)} = \frac{\text{their}14.8}{\sin 35}\) oe | M1 | |
| 25.69 to 25.8 | A1 | Allow B2 for \(25.69 \leq AD < 25.8\) unsupported; B0 for 25.8 unsupported; M0A0 for \(\frac{14.8}{\cos 55} = 25.803...\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Area of \(ABC = 48\) soi | B1 | 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. Condone 48.0... |
| \(\frac{1}{2} \times \text{their}14.8 \times \text{their}25.7 \times \sin(\text{their}59.6 - 10)\) | M1 | |
| 192.8 to 194 \([\text{m}^2]\) | A1 | B3 for correct answer in range if unsupported |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Angle \(HMG = \frac{\pi - 1.1}{2}\) or \(MHG = 0.55\) \((31.5126°)\) | B1 | or angle \(EMF\) or angle \(MEF\); allow 1.02 to 1.021 or 58.487° to 58.5° |
| \(HM = 1.7176\) to \(1.7225\) | B1 | May be implied by final answer |
| \(\frac{1}{2} \times 1.1 \times \text{their } HM^2\) or \(\frac{\theta}{360} \times \pi \times \text{their} HM^2\) | M1 | \(1.63(0661924...)\); \(\theta = 63(.025357...)\); check arithmetic if necessary; their \(HM \neq 0.9\) or \(1.8\) |
| Area of triangle \(EMF = 0.652\) to \(0.662\) | B1 | or \(MGH\); may be implied by final answer or in double this (1.304 to 1.324) |
| 2.95 to 2.952 \([\text{m}^2]\) cao | A1 | Full marks may be awarded for final answer in correct range ie allow recovery of accuracy |
## Question 2(i)(A):
| Answer | Mark | Guidance |
|--------|------|----------|
| $AC^2 = 12.8^2 + 7.5^2$ oe | M1 | Allow correct application of cosine rule or from finding relevant angle and using trig |
| $AC = 14.83543056...$ | A1 | Rot to 3 or more sf, or 15. **B2** for 14.8 or better unsupported |
| $\tan C = \frac{12.8}{7.5}$ or $C = 90 - \tan^{-1}\left(\frac{7.5}{12.8}\right)$ oe | M1 | or $\sin C = \frac{12.8}{\text{their}14.8}$ or $\cos C = \frac{7.5}{\text{their}14.8}$; or $\frac{\sin C}{12.8} = \frac{\sin 90}{\text{their}14.8}$; or $\cos C = \frac{\text{their}14.8^2 + 7.5^2 - 12.8^2}{2 \times 7.5 \times \text{their}14.8}$ |
| 59.6 to 59.64 | A1 | |
| $\frac{AD}{\sin(155 - \text{their}59.6)} = \frac{\text{their}14.8}{\sin 35}$ oe | M1 | |
| 25.69 to 25.8 | A1 | Allow **B2** for $25.69 \leq AD < 25.8$ unsupported; **B0** for 25.8 unsupported; **M0A0** for $\frac{14.8}{\cos 55} = 25.803...$ |
---
## Question 2(i)(B):
| Answer | Mark | Guidance |
|--------|------|----------|
| Area of $ABC = 48$ soi | B1 | 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. Condone 48.0... |
| $\frac{1}{2} \times \text{their}14.8 \times \text{their}25.7 \times \sin(\text{their}59.6 - 10)$ | M1 | |
| 192.8 to 194 $[\text{m}^2]$ | A1 | **B3** for correct answer in range if unsupported |
---
## Question 2(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Angle $HMG = \frac{\pi - 1.1}{2}$ or $MHG = 0.55$ $(31.5126°)$ | B1 | or angle $EMF$ or angle $MEF$; allow 1.02 to 1.021 or 58.487° to 58.5° |
| $HM = 1.7176$ to $1.7225$ | B1 | May be implied by final answer |
| $\frac{1}{2} \times 1.1 \times \text{their } HM^2$ or $\frac{\theta}{360} \times \pi \times \text{their} HM^2$ | M1 | $1.63(0661924...)$; $\theta = 63(.025357...)$; check arithmetic if necessary; their $HM \neq 0.9$ or $1.8$ |
| Area of triangle $EMF = 0.652$ to $0.662$ | B1 | or $MGH$; may be implied by final answer or in double this (1.304 to 1.324) |
| 2.95 to 2.952 $[\text{m}^2]$ cao | A1 | Full marks may be awarded for final answer in correct range ie allow recovery of accuracy |
---2 Fig. 10.1 shows Jean's back garden. This is a quadrilateral ABCD with dimensions as shown.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{cd507fa5-97b2-4edd-ae37-a58aea1de5ed-2_711_1018_292_549}
\captionsetup{labelformat=empty}
\caption{Fig. 10.1}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item (A) Calculate AC and angle ACB . Hence calculate AD .\\
(B) Calculate the area of the garden.
\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. $\mathrm { FG } = 1.8 \mathrm {~m}$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{cd507fa5-97b2-4edd-ae37-a58aea1de5ed-2_579_981_1512_567}
\captionsetup{labelformat=empty}
\caption{Fig. 10.2}
\end{center}
\end{figure}
Calculate the area of one of these fence panels.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C2 Q2 [14]}}