| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2018 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Bearings and navigation |
| Difficulty | Moderate -0.3 This is a standard bearings question requiring angle calculation from bearings (part a), cosine rule application (part b), and sine rule with bearing conversion (part c). All techniques are routine for C2 level, though the multi-step nature and bearing conversions require care. Slightly easier than average due to straightforward application of standard methods with no novel problem-solving 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/Working | Mark | Guidance |
| \((APN =) 360° - 314° = 46°\); \((APB =) 46° + 52° = 98°\) or (Reflex \(APB\)) \(= 314° - 52° = 262°\); \((APB =) 360° - 262° = 98°\) | B1 | Correct explanation that explains why \(APN\) is \(46°\) (e.g. \(360°-314°\)) and adds \(52°\), or shows reflex \(APB = 262°\) so \(APB = 360°-262°=98°\). Do not allow use of \(AB = 9.8\) from (b) |
| Total: 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((AB^2 =) 8.7^2 + 3.5^2 - 2 \times 8.7 \times 3.5\cos 98°\) | M1 | Correct use of cosine rule. Look for \(8.7^2 + 3.5^2 - 2\times8.7\times3.5\cos98°\) |
| \(AB = 9.8\) (km) | A1 | Awrt 9.8 km |
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{\text{"9.8"}}{\sin 98°} = \frac{3.5}{\sin PAB}\) or \(3.5^2 = 8.7^2 + \text{"9.8"}^2 - 2\times8.7\times\text{"9.8"}\cos PAB\) | M1 | Correct sine or cosine rule method to obtain angle \(PAB\). May be implied by awrt \(21°\) |
| \(PAB = 20.66...°\) | A1 | Awrt \(21°\). May be implied by a correct bearing |
| Bearing is \(180° - \text{"20.66°"} - 46°\) | M1 | Fully correct method |
| \(= 113°\) or \(114°\) | A1 | Awrt \(113°\) or awrt \(114°\) |
| Total: 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{\text{"9.8"}}{\sin 98°} = \frac{8.7}{\sin PBA}\) or \(8.7^2 = 3.5^2 + \text{"9.8"}^2 - 2\times3.5\times\text{"9.8"}\cos PBA\) | M1 | Correct sine or cosine rule to obtain angle \(PBA\). May be implied by awrt \(61°\) or \(62°\) |
| \(PBA = 61.33...°\) | A1 | Awrt \(61°\) or \(62°\). May be implied by a correct bearing |
| Bearing is \(52° + \text{"61.33...°"}\) | M1 | Fully correct method |
| \(= 113°\) or \(114°\) | A1 | Awrt \(113°\) or awrt \(114°\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let \(\alpha =\) Bearing \(- 90°\); \(\tan\alpha = \frac{BC}{AC} = \frac{8.7\cos46° - 3.5\cos52°}{8.7\sin46° + 3.5\sin52°}\) | M1 | Correct method for \(\alpha\) |
| \(\alpha = 23.33°\) | A1 | Awrt \(23°\). May be implied by a correct bearing |
| Bearing is \(90° + \text{"23.33°"}\) | M1 | Fully correct method |
| \(= 113°\) or \(114°\) | A1 | Awrt \(113°\) or awrt \(114°\) |
| Total: 7 marks |
## Question 13:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(APN =) 360° - 314° = 46°$; $(APB =) 46° + 52° = 98°$ or (Reflex $APB$) $= 314° - 52° = 262°$; $(APB =) 360° - 262° = 98°$ | B1 | Correct explanation that explains why $APN$ is $46°$ (e.g. $360°-314°$) and adds $52°$, or shows reflex $APB = 262°$ so $APB = 360°-262°=98°$. Do not allow use of $AB = 9.8$ from (b) |
| **Total: 1 mark** | | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(AB^2 =) 8.7^2 + 3.5^2 - 2 \times 8.7 \times 3.5\cos 98°$ | M1 | Correct use of cosine rule. Look for $8.7^2 + 3.5^2 - 2\times8.7\times3.5\cos98°$ |
| $AB = 9.8$ (km) | A1 | Awrt 9.8 km |
| **Total: 2 marks** | | |
### Part (c) - Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{\text{"9.8"}}{\sin 98°} = \frac{3.5}{\sin PAB}$ or $3.5^2 = 8.7^2 + \text{"9.8"}^2 - 2\times8.7\times\text{"9.8"}\cos PAB$ | M1 | Correct sine or cosine rule method to obtain angle $PAB$. May be implied by awrt $21°$ |
| $PAB = 20.66...°$ | A1 | Awrt $21°$. May be implied by a correct bearing |
| Bearing is $180° - \text{"20.66°"} - 46°$ | M1 | Fully correct method |
| $= 113°$ or $114°$ | A1 | Awrt $113°$ or awrt $114°$ |
| **Total: 4 marks** | | |
### Part (c) - Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{\text{"9.8"}}{\sin 98°} = \frac{8.7}{\sin PBA}$ or $8.7^2 = 3.5^2 + \text{"9.8"}^2 - 2\times3.5\times\text{"9.8"}\cos PBA$ | M1 | Correct sine or cosine rule to obtain angle $PBA$. May be implied by awrt $61°$ or $62°$ |
| $PBA = 61.33...°$ | A1 | Awrt $61°$ or $62°$. May be implied by a correct bearing |
| Bearing is $52° + \text{"61.33...°"}$ | M1 | Fully correct method |
| $= 113°$ or $114°$ | A1 | Awrt $113°$ or awrt $114°$ |
### Part (c) - Way 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $\alpha =$ Bearing $- 90°$; $\tan\alpha = \frac{BC}{AC} = \frac{8.7\cos46° - 3.5\cos52°}{8.7\sin46° + 3.5\sin52°}$ | M1 | Correct method for $\alpha$ |
| $\alpha = 23.33°$ | A1 | Awrt $23°$. May be implied by a correct bearing |
| Bearing is $90° + \text{"23.33°"}$ | M1 | Fully correct method |
| $= 113°$ or $114°$ | A1 | Awrt $113°$ or awrt $114°$ |
| **Total: 7 marks** | | |13.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6f9ace43-747b-419f-a9d1-d30165d77379-42_840_1010_287_571}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows the position of two stationary boats, $A$ and $B$, and a port $P$ which are assumed to be in the same horizontal plane.
Boat $A$ is 8.7 km on a bearing of $314 ^ { \circ }$ from port $P$.\\
Boat $B$ is 3.5 km on a bearing of $052 ^ { \circ }$ from port $P$.
\begin{enumerate}[label=(\alph*)]
\item Show that angle $A P B$ is $98 ^ { \circ }$
\item Find the distance of boat $B$ from boat $A$, giving your answer to one decimal place.
\item Find the bearing of boat $B$ from boat $A$, giving your answer to the nearest degree.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2018 Q13 [7]}}