| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Real-world application problems |
| Difficulty | Standard +0.3 This is a straightforward application of sine/cosine rules and basic circle geometry. Part (i) uses cosine rule then bearing calculation, (ii) is standard triangle area, (iii) involves arc length to find angle and sector area subtraction. All steps are routine with clear signposting and 'show that' answers provided, 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)1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| Using cosine rule: \(\cos(\angle ACB) = \frac{118^2 + 82^2 - 189^2}{2(118)(82)}\) | M1 | Correct formula |
| \(= \frac{13924 + 6724 - 35721}{19352} = \frac{-15073}{19352}\) | A1 | Correct substitution |
| \(\angle ACB = 141.1°\) | A1 | |
| Bearing \(= 180° - 141.1° = 038.9°\) (approx) | M1 A1 | Correct bearing logic; \(038.9°\) |
| Answer | Marks |
|---|---|
| Area \(= \frac{1}{2}(118)(82)\sin(\angle ACB)\) | M1 |
| \(= \frac{1}{2}(118)(82)\sin(141.1°) \approx 3040\text{ m}^2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(OA = OB = 130\), \(AB = 189\) | ||
| \(\cos(\angle AOB) = \frac{130^2 + 130^2 - 189^2}{2(130)(130)}\) | M1 | Cosine rule |
| \(= \frac{33800 - 35721}{33800} = -0.05685\) | A1 | |
| \(\angle AOB = 1.627... \approx 1.63\) radians | A1 | Shown correctly |
| Answer | Marks | Guidance |
|---|---|---|
| Area of sector \(= \frac{1}{2}(130^2)(1.63) = 13773.5\) | M1 | Correct sector formula |
| Area of triangle \(OAB = \frac{1}{2}(130^2)\sin(1.63) = 8450\) | M1 | |
| Area of original triangle \(\approx 3040\) | ||
| Added area \(=\) sector \(-\) triangle \(OAB +\) triangle \(ACB\) or equivalent method \(\approx 5300\text{ m}^2\) | M1 A1 | Correct method shown; 5300 to 2 s.f. |
## Question 11(i):
Using cosine rule: $\cos(\angle ACB) = \frac{118^2 + 82^2 - 189^2}{2(118)(82)}$ | M1 | Correct formula
$= \frac{13924 + 6724 - 35721}{19352} = \frac{-15073}{19352}$ | A1 | Correct substitution
$\angle ACB = 141.1°$ | A1 |
Bearing $= 180° - 141.1° = 038.9°$ (approx) | M1 A1 | Correct bearing logic; $038.9°$
## Question 11(ii):
Area $= \frac{1}{2}(118)(82)\sin(\angle ACB)$ | M1 |
$= \frac{1}{2}(118)(82)\sin(141.1°) \approx 3040\text{ m}^2$ | A1 |
## Question 11(iii)(A):
$OA = OB = 130$, $AB = 189$ |
$\cos(\angle AOB) = \frac{130^2 + 130^2 - 189^2}{2(130)(130)}$ | M1 | Cosine rule
$= \frac{33800 - 35721}{33800} = -0.05685$ | A1 |
$\angle AOB = 1.627... \approx 1.63$ radians | A1 | Shown correctly
## Question 11(iii)(B):
Area of sector $= \frac{1}{2}(130^2)(1.63) = 13773.5$ | M1 | Correct sector formula
Area of triangle $OAB = \frac{1}{2}(130^2)\sin(1.63) = 8450$ | M1 |
Area of original triangle $\approx 3040$ |
Added area $=$ sector $-$ triangle $OAB +$ triangle $ACB$ or equivalent method $\approx 5300\text{ m}^2$ | M1 A1 | Correct method shown; 5300 to 2 s.f.
---11 Fig. 11.1 shows a village green which is bordered by 3 straight roads $A B , B C$ and $C A$. The road AC runs due North and the measurements shown are in metres.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b4c0b4b0-f13c-49a9-9f98-f86f28d1f577-4_460_1143_486_591}
\captionsetup{labelformat=empty}
\caption{Fig. 11.1}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Calculate the bearing of B from C , giving your answer to the nearest $0.1 ^ { \circ }$.
\item Calculate the area of the village green.
The road AB is replaced by a new road, as shown in Fig. 11.2. The village green is extended up to the new road.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b4c0b4b0-f13c-49a9-9f98-f86f28d1f577-4_440_1002_1436_737}
\captionsetup{labelformat=empty}
\caption{Fig. 11.2}
\end{center}
\end{figure}
The new road is an arc of a circle with centre O and radius 130 m .
\item (A) Show that angle AOB is 1.63 radians, correct to 3 significant figures.\\
(B) Show that the area of land added to the village green is $5300 \mathrm {~m} ^ { 2 }$ correct to 2 significant figures.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C2 2007 Q11 [12]}}