| 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 | Standard +0.3 This is a straightforward application of the sine rule in triangle ABD to find angle BAD, followed by using the cosine rule in triangle BCD to find BC. Both are standard two-step calculations requiring routine application of formulae with no novel insight, making it slightly easier than average. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{\sin A}{5.6} = \frac{\sin 79}{8.4}\) | M1 | |
| \([A =]\) 40.87 to 41 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([BC^2 =] 5.6^2 + 7.8^2 - 2 \times 5.6 \times 7.8 \times \cos(180 - 79°)\) | M1 | |
| \(= 108.8\) to \(108.9\) | A1 | |
| \([BC =] 10.4(...)\) | A1 |
## Question 6(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{\sin A}{5.6} = \frac{\sin 79}{8.4}$ | M1 | |
| $[A =]$ 40.87 to 41 | A1 | |
---
## Question 6(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[BC^2 =] 5.6^2 + 7.8^2 - 2 \times 5.6 \times 7.8 \times \cos(180 - 79°)$ | M1 | |
| $= 108.8$ to $108.9$ | A1 | |
| $[BC =] 10.4(...)$ | A1 | |6\\
Not to scale
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{cd507fa5-97b2-4edd-ae37-a58aea1de5ed-5_484_968_1516_617}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{center}
\end{figure}
Fig. 7 shows triangle ABC , with $\mathrm { AB } = 8.4 \mathrm {~cm}$. D is a point on AC such that angle $\mathrm { ADB } = 79 ^ { \circ }$, $\mathrm { BD } = 5.6 \mathrm {~cm}$ and $\mathrm { CD } = 7.8 \mathrm {~cm}$.
Calculate\\
(i) angle BAD ,\\
(ii) the length BC .
\hfill \mbox{\textit{OCR MEI C2 Q6 [5]}}