| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2014 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Bearings and navigation |
| Difficulty | Moderate -0.8 This is a straightforward application of the cosine rule followed by the sine rule in a bearings context. The angle at B is easily found (90° + 65° = 155°), then standard two-step calculation with clear diagram interpretation. Slightly easier than average due to being a routine textbook-style bearings problem with no tricky angle work or multi-stage reasoning. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.05b Sine and cosine rules: including ambiguous case |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(AC^2 = 10^2 + 8^2 - 2 \times 10 \times 8\cos65°\Rightarrow AC = ..\) | M1 | Uses cosine rule to find \(AC\). Rule if stated must be correct. If not stated must be correct form. Alternative methods (e.g. dropping perpendicular) must be fully complete. |
| \(AC = 9.8...\) | A1 | Accept answers rounding or truncating to \(AC = 9.8...\)km |
| \(AC = 9.82\)km (9820m) (to nearest 10m) | A1 | Accept \(9.82\)km or \(9820\)m. Both accuracy and units necessary. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{\sin A}{8} = \frac{\sin65°}{'9.817..'} \Rightarrow A=\) OR \(\frac{\sin C}{10} = \frac{\sin65°}{'9.817..'} \Rightarrow C=\) | M1 | Uses sine rule (or cosine rule) with their \(AC\) to find angle \(A\) or \(C\). Sides and angles must be correctly matched. |
| \(\angle A =\) awrt \(47.6°\) or \(\angle C =\) awrt \(67.4°\) | A1 | Don't be overly concerned with labelling of angle. |
| Bearing \(=\) awrt \(132.4°\) | A1ft | Follow through on \((180-A)°\) if found \(A\), or \((65+C)°\) if found \(C\). |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $AC^2 = 10^2 + 8^2 - 2 \times 10 \times 8\cos65°\Rightarrow AC = ..$ | M1 | Uses cosine rule to find $AC$. Rule if stated must be correct. If not stated must be correct form. Alternative methods (e.g. dropping perpendicular) must be fully complete. |
| $AC = 9.8...$ | A1 | Accept answers rounding or truncating to $AC = 9.8...$km |
| $AC = 9.82$km (9820m) (to nearest 10m) | A1 | Accept $9.82$km or $9820$m. Both accuracy and units necessary. |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{\sin A}{8} = \frac{\sin65°}{'9.817..'} \Rightarrow A=$ OR $\frac{\sin C}{10} = \frac{\sin65°}{'9.817..'} \Rightarrow C=$ | M1 | Uses sine rule (or cosine rule) with their $AC$ to find angle $A$ or $C$. Sides and angles must be correctly matched. |
| $\angle A =$ awrt $47.6°$ or $\angle C =$ awrt $67.4°$ | A1 | Don't be overly concerned with labelling of angle. |
| Bearing $=$ awrt $132.4°$ | A1ft | Follow through on $(180-A)°$ if found $A$, or $(65+C)°$ if found $C$. |
---1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b85872d4-00b2-499b-9765-f7559d3de66a-02_856_700_214_630}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows the position of three stationary fishing boats $A , B$ and $C$, which are assumed to be in the same horizontal plane.
Boat $A$ is 10 km due north of boat $B$.
Boat $C$ is 8 km on a bearing of $065 ^ { \circ }$ from boat $B$.
\begin{enumerate}[label=(\alph*)]
\item Find the distance of boat $C$ from boat $A$, giving your answer to the nearest 10 metres.
\item Find the bearing of boat $C$ from boat $A$, giving your answer to one decimal place.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2014 Q1 [6]}}