| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Triangle area calculation |
| Difficulty | Moderate -0.3 Part (a) is a standard application of Heron's formula or cosine rule followed by area formula, requiring straightforward calculation with given side lengths. Part (b) adds a contextual reasoning element about measurement accuracy, which is slightly unusual but accessible. Overall, this is slightly easier than average due to being a direct application of standard formulas with no geometric insight required. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cos A = \frac{100^2 + 120^2 - 135^2}{2 \times 100 \times 120}\) | M1 | 3.1a — Alt: \(\cos B = \frac{100^2 + 135^2 - 120^2}{2 \times 100 \times 135}\) OR \(\cos C = \frac{120^2 + 135^2 - 100^2}{2 \times 120 \times 135}\) |
| \(\cos A = 0.2572916...\) | A1 | 1.1 — Alt: \(\cos B = 0.512037...\) OR \(\cos C = 0.698302...\) |
| \([A =]\ 75.09058...\) | A1 | 1.1 — May be implied; \(B = 59.200...\) OR \(C = 45.7090...\) |
| Area \(= \frac{1}{2} \times 100 \times 120 \times \sin(\text{their } A)\) | M1 | 3.1a — Alt: Area \(= \frac{1}{2} \times 100 \times 135 \times \sin(\text{their } B)\) OR \(\frac{1}{2} \times 120 \times 135 \times \sin(\text{their } C)\) |
| \(5800\ [\text{m}^2]\) | A1 | 1.1 — Accept answers to greater degree of accuracy |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| E.g. The sides might only be to the nearest 5 metres so the possible areas cover quite a big range. E.g. The sides are no more accurate than to the nearest metre, so could be half a metre out. Taking half a metre off each side would lose more than \(1\ \text{m}^2\) of area | E1 | 3.2b — Correct explanation |
| [1] |
# Question 5:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos A = \frac{100^2 + 120^2 - 135^2}{2 \times 100 \times 120}$ | M1 | 3.1a — Alt: $\cos B = \frac{100^2 + 135^2 - 120^2}{2 \times 100 \times 135}$ OR $\cos C = \frac{120^2 + 135^2 - 100^2}{2 \times 120 \times 135}$ |
| $\cos A = 0.2572916...$ | A1 | 1.1 — Alt: $\cos B = 0.512037...$ OR $\cos C = 0.698302...$ |
| $[A =]\ 75.09058...$ | A1 | 1.1 — May be implied; $B = 59.200...$ OR $C = 45.7090...$ |
| Area $= \frac{1}{2} \times 100 \times 120 \times \sin(\text{their } A)$ | M1 | 3.1a — Alt: Area $= \frac{1}{2} \times 100 \times 135 \times \sin(\text{their } B)$ OR $\frac{1}{2} \times 120 \times 135 \times \sin(\text{their } C)$ |
| $5800\ [\text{m}^2]$ | A1 | 1.1 — Accept answers to greater degree of accuracy |
| **[5]** | | |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| E.g. The sides might only be to the nearest 5 metres so the possible areas cover quite a big range. E.g. The sides are no more accurate than to the nearest metre, so could be half a metre out. Taking half a metre off each side would lose more than $1\ \text{m}^2$ of area | E1 | 3.2b — Correct explanation |
| **[1]** | | |
---5 A triangular field has sides of length $100 \mathrm {~m} , 120 \mathrm {~m}$ and 135 m .
\begin{enumerate}[label=(\alph*)]
\item Find the area of the field.
\item Explain why it would not be reasonable to expect your answer in (a) to be accurate to the nearest square metre.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 2 Q5 [6]}}