| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Radians, Arc Length and Sector Area |
| Type | Segment area calculation |
| Difficulty | Standard +0.3 This is a straightforward multi-part question combining basic trigonometry (sine rule for part i), angle calculation using congruent triangles (part ii(a)), and standard sector area formula application (part ii(b)). All techniques are routine C2 content with clear scaffolding and one answer given to guide students. Slightly above average only due to the multi-step nature and visualization required. |
| 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 |
|---|---|---|
| \(\frac{\sin\theta}{8} = \frac{\sin 65}{11}\) | M1 | Attempt use of correct sine rule |
| \(\theta = 41.2°\) | A1 | Obtain \(41.2°\), or better |
| Answer | Marks | Guidance |
|---|---|---|
| \(180 - (2 \times 65) = 50°\) or \(65 \times \frac{\pi}{180} = 1.134\) | M1 | Use conversion factor of \(\frac{\pi}{180}\) |
| \(50 \times \frac{\pi}{180} = 0.873\) A.G. \(\pi - (2 \times 1.134) = 0.873\) | A1 | Show 0.873 radians convincingly (AG) |
| Answer | Marks | Guidance |
|---|---|---|
| area sector \(= \frac{1}{2} \times 8^2 \times 0.873 = 27.9\) | M1 | Attempt area of sector, using \(\frac{1}{2}r^2\theta\) |
| area triangle \(= \frac{1}{2} \times 8^2 \times \sin 0.873 = 24.5\) | M1 | Attempt area of triangle using \(\frac{1}{2}r^2\sin\theta\) |
| area segment \(= 27.9 - 24.5 = 3.41\) | M1 | Subtract area of triangle from area of sector |
| A1 | Obtain 3.41 or 3.42 |
## Question 5:
### Part (i)
| $\frac{\sin\theta}{8} = \frac{\sin 65}{11}$ | M1 | Attempt use of correct sine rule |
| $\theta = 41.2°$ | A1 | Obtain $41.2°$, or better |
### Part (ii)(a)
| $180 - (2 \times 65) = 50°$ or $65 \times \frac{\pi}{180} = 1.134$ | M1 | Use conversion factor of $\frac{\pi}{180}$ |
| $50 \times \frac{\pi}{180} = 0.873$ **A.G.** $\pi - (2 \times 1.134) = 0.873$ | A1 | Show 0.873 radians convincingly **(AG)** |
### Part (ii)(b)
| area sector $= \frac{1}{2} \times 8^2 \times 0.873 = 27.9$ | M1 | Attempt area of sector, using $\frac{1}{2}r^2\theta$ |
| area triangle $= \frac{1}{2} \times 8^2 \times \sin 0.873 = 24.5$ | M1 | Attempt area of triangle using $\frac{1}{2}r^2\sin\theta$ |
| area segment $= 27.9 - 24.5 = 3.41$ | M1 | Subtract area of triangle from area of sector |
| | A1 | Obtain 3.41 or 3.42 |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{570435e0-5685-4c5b-9ed8-f2bc22bdfb24-02_396_1070_1768_536}
The diagram shows two congruent triangles, $B C D$ and $B A E$, where $A B C$ is a straight line. In triangle $B C D , B D = 8 \mathrm {~cm} , C D = 11 \mathrm {~cm}$ and angle $C B D = 65 ^ { \circ }$. The points $E$ and $D$ are joined by an arc of a circle with centre $B$ and radius 8 cm .
\begin{enumerate}[label=(\roman*)]
\item Find angle $B C D$.
\item (a) Show that angle $E B D$ is 0.873 radians, correct to 3 significant figures.\\
(b) Hence find the area of the shaded segment bounded by the chord $E D$ and the arc $E D$, giving your answer correct to 3 significant figures.
\end{enumerate}
\hfill \mbox{\textit{OCR C2 2010 Q5 [8]}}