| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Point on side of triangle |
| Difficulty | Moderate -0.3 This is a straightforward two-part application of the cosine rule and triangle area formula. Part (i) requires recognizing that angle ACD = 53.4° gives angle ACB = 180° - 53.4° = 126.6° (supplementary angles on straight line), then applying cosine rule in triangle ACD. Part (ii) uses the standard area formula (1/2)ab sin C. Both parts are routine calculations with no problem-solving insight 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) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(9.8^2 + 6.4^2 - 2 \times 9.8 \times 6.4 \times \cos 53.4\) | M1 | |
| \(9.8^2 + 6.4^2 - 74.79... [= 62.2...]\) | M1 | For evidence of correct order of operations used; may be implied by correct answer. 6.89 implies M0; 262.4368 implies M1 (calc in radian mode); NB \(9.8\sin 53.4 = 7.87\) |
| 7.887... or 7.89 or 7.9 | A1 | If M0, B3 for 7.89 or more precise www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{2} \times 9.8 \times 7.3 \times \sin(180 - 53.4)\) oe seen | M1 | or \(\sin 53.4\) used; may be embedded; may be split into height \(= 9.8 \times \sin 53.4\) then Area \(= \frac{1}{2} \times 7.3 \times \text{height}\) |
| 28.716... or 28.72 or 28.7 or 29 isw | A1 | If M0, B2 for 28.7 or more precise www |
## Question 3(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $9.8^2 + 6.4^2 - 2 \times 9.8 \times 6.4 \times \cos 53.4$ | M1 | |
| $9.8^2 + 6.4^2 - 74.79... [= 62.2...]$ | M1 | For evidence of correct order of operations used; may be implied by correct answer. 6.89 implies M0; 262.4368 implies M1 (calc in radian mode); NB $9.8\sin 53.4 = 7.87$ |
| 7.887... or 7.89 or 7.9 | A1 | If M0, B3 for 7.89 or more precise www |
---
## Question 3(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{2} \times 9.8 \times 7.3 \times \sin(180 - 53.4)$ oe seen | M1 | or $\sin 53.4$ used; may be embedded; may be split into height $= 9.8 \times \sin 53.4$ then Area $= \frac{1}{2} \times 7.3 \times \text{height}$ |
| 28.716... or 28.72 or 28.7 or 29 isw | A1 | If M0, B2 for 28.7 or more precise www |
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\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{cd507fa5-97b2-4edd-ae37-a58aea1de5ed-3_596_689_244_534}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
\section*{Not to scale}
In Fig. 3, BCD is a straight line. $\mathrm { AC } = 9.8 \mathrm {~cm} , \mathrm { BC } = 7.3 \mathrm {~cm}$ and $\mathrm { CD } = 6.4 \mathrm {~cm}$; angle $\mathrm { ACD } = 53.4 ^ { \circ }$.\\
(i) Calculate the length AD .\\
(ii) Calculate the area of triangle ABC .
\hfill \mbox{\textit{OCR MEI C2 Q3 [5]}}