OCR C4 — Question 3 19 marks

Exam BoardOCR
ModuleC4 (Core Mathematics 4)
Marks19
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRadians, Arc Length and Sector Area
TypeExact form answers
DifficultyChallenging +1.2 This is a structured multi-part question requiring trigonometric identities, exact value manipulation, and geometric reasoning. While it involves several steps and nested radicals, each part is clearly signposted with standard techniques (double angle formulas, solving quadratics). The historical context and geometric setup add interest but don't significantly increase difficulty. Slightly above average due to length and exact form manipulation, but well within reach of a competent C4 student.
Spec1.05g Exact trigonometric values: for standard angles1.05l Double angle formulae: and compound angle formulae
3 Archimedes, about 2200 years ago, used regular polygons inside and outside circles to obtain approximations for \(\pi\).
  1. Fig. 8.1 shows a regular 12 -sided polygon inscribed in a circle of radius 1 unit, centre \(\mathrm { O } . \mathrm { AB }\) is one of the sides of the polygon. C is the midpoint of AB . Archimedes used the fact that the circumference of the circle is greater than the perimeter of this polygon. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{c0fcd64b-8ca0-4309-9f58-c23cc4208f4d-2_457_422_457_936} \captionsetup{labelformat=empty} \caption{Fig. 8.1}
    \end{figure} (A) Show that \(\mathrm { AB } = 2 \sin 15 ^ { \circ }\).
    (B) Use a double angle formula to express \(\cos 30 ^ { \circ }\) in terms of \(\sin 15 ^ { \circ }\). Using the exact value of \(\cos 30 ^ { \circ }\), show that \(\sin 15 ^ { \circ } = \frac { 1 } { 2 } \sqrt { 2 - \sqrt { 3 } }\).
    (C) Use this result to find an exact expression for the perimeter of the polygon. Hence show that \(\pi > 6 \sqrt { 2 - \sqrt { 3 } }\).
  2. In Fig. 8.2, a regular 12-sided polygon lies outside the circle of radius 1 unit, which touches each side of the polygon. F is the midpoint of DE. Archimedes used the fact that the circumference of the circle is less than the perimeter of this polygon. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{c0fcd64b-8ca0-4309-9f58-c23cc4208f4d-2_450_416_1562_940} \captionsetup{labelformat=empty} \caption{Fig. 8.2}
    \end{figure} (A) Show that \(\mathrm { DE } = 2 \tan 15 ^ { \circ }\).
    (B) Let \(t = \tan 15 ^ { \circ }\). Use a double angle formula to express \(\tan 30 ^ { \circ }\) in terms of \(t\). Hence show that \(t ^ { 2 } + 2 \sqrt { 3 } t - 1 = 0\).
    (C) Solve this equation, and hence show that \(\pi < 12 ( 2 - \sqrt { 3 } )\).
  3. Use the results in parts (i)( \(C\) ) and (ii)( \(C\) ) to establish upper and lower bounds for the value of \(\pi\), giving your answers in decimal form.
Question 3:
Part (i)(A):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(360° \div 24 = 15°\); \(CB/OB = \sin 15°\)M1 \(AB = 2AC\) or \(2CB\); \(\angle AOC = 15°\) o.e.
\(CB = 1\sin 15°\)E1
\(AB = 2CB = 2\sin 15°\)*[2]
Part (i)(B):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\cos 30° = 1 - 2\sin^2 15°\)B1
\(\cos 30° = \frac{\sqrt{3}}{2}\)B1
\(\frac{\sqrt{3}}{2} = 1 - 2\sin^2 15°\)
\(2\sin^2 15° = 1 - \frac{\sqrt{3}}{2} = \frac{(2-\sqrt{3})}{2}\)M1 simplifying
\(\sin^2 15° = \frac{(2-\sqrt{3})}{4}\)
\(\sin 15° = \sqrt{\frac{2-\sqrt{3}}{4}} = \frac{1}{2}\sqrt{2-\sqrt{3}}\)*E1 [4]
Part (i)(C):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Perimeter \(= 12 \times AB = 24 \times \frac{1}{2}\sqrt{(2-\sqrt{3})} = 12\sqrt{(2-\sqrt{3})}\)M1
circumference of circle \(>\) perimeter of polygon
\(\Rightarrow 2\pi > 12\sqrt{(2-\sqrt{3})}\)E1
\(\Rightarrow \pi > 6\sqrt{(2-\sqrt{3})}\)*[2]
Part (ii)(A):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\tan 15° = FE/OF\)M1
\(FE = \tan 15°\)E1
\(DE = 2FE = 2\tan 15°\)*[2]
Part (ii)(B):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\tan 30° = \frac{2\tan 15}{1-\tan^2 15} = \frac{2t}{1-t^2}\)B1
\(\tan 30° = \frac{1}{\sqrt{3}}\)
\(\Rightarrow \frac{2t}{1-t^2} = \frac{1}{\sqrt{3}} \Rightarrow 2\sqrt{3}t = 1-t^2\)M1
\(\Rightarrow t^2 + 2\sqrt{3}t - 1 = 0\)*E1 [3]
Part (ii)(C):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(t = \frac{-2\sqrt{3} \pm \sqrt{12+4}}{2} = 2 - \sqrt{3}\)M1 A1 using positive root from exact working
circumference \(<\) perimeter
\(\Rightarrow 2\pi < 24(2-\sqrt{3})\)M1
\(\Rightarrow \pi < 12(2-\sqrt{3})\)*E1 [4]
Part (iii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(6\sqrt{(2-\sqrt{3})} < \pi < 12(2-\sqrt{3})\)
\(\Rightarrow 3.106 < \pi < 3.215\)B1 B1 [2] 3.106, 3.215 
## Question 3:

### Part (i)(A):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $360° \div 24 = 15°$; $CB/OB = \sin 15°$ | M1 | $AB = 2AC$ or $2CB$; $\angle AOC = 15°$ o.e. |
| $CB = 1\sin 15°$ | E1 | |
| $AB = 2CB = 2\sin 15°$* | [2] | |

### Part (i)(B):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\cos 30° = 1 - 2\sin^2 15°$ | B1 | |
| $\cos 30° = \frac{\sqrt{3}}{2}$ | B1 | |
| $\frac{\sqrt{3}}{2} = 1 - 2\sin^2 15°$ | | |
| $2\sin^2 15° = 1 - \frac{\sqrt{3}}{2} = \frac{(2-\sqrt{3})}{2}$ | M1 | simplifying |
| $\sin^2 15° = \frac{(2-\sqrt{3})}{4}$ | | |
| $\sin 15° = \sqrt{\frac{2-\sqrt{3}}{4}} = \frac{1}{2}\sqrt{2-\sqrt{3}}$* | E1 [4] | |

### Part (i)(C):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Perimeter $= 12 \times AB = 24 \times \frac{1}{2}\sqrt{(2-\sqrt{3})} = 12\sqrt{(2-\sqrt{3})}$ | M1 | |
| circumference of circle $>$ perimeter of polygon | | |
| $\Rightarrow 2\pi > 12\sqrt{(2-\sqrt{3})}$ | E1 | |
| $\Rightarrow \pi > 6\sqrt{(2-\sqrt{3})}$* | [2] | |

### Part (ii)(A):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan 15° = FE/OF$ | M1 | |
| $FE = \tan 15°$ | E1 | |
| $DE = 2FE = 2\tan 15°$* | [2] | |

### Part (ii)(B):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan 30° = \frac{2\tan 15}{1-\tan^2 15} = \frac{2t}{1-t^2}$ | B1 | |
| $\tan 30° = \frac{1}{\sqrt{3}}$ | | |
| $\Rightarrow \frac{2t}{1-t^2} = \frac{1}{\sqrt{3}} \Rightarrow 2\sqrt{3}t = 1-t^2$ | M1 | |
| $\Rightarrow t^2 + 2\sqrt{3}t - 1 = 0$* | E1 [3] | |

### Part (ii)(C):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $t = \frac{-2\sqrt{3} \pm \sqrt{12+4}}{2} = 2 - \sqrt{3}$ | M1 A1 | using positive root from exact working |
| circumference $<$ perimeter | | |
| $\Rightarrow 2\pi < 24(2-\sqrt{3})$ | M1 | |
| $\Rightarrow \pi < 12(2-\sqrt{3})$* | E1 [4] | |

### Part (iii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $6\sqrt{(2-\sqrt{3})} < \pi < 12(2-\sqrt{3})$ | | |
| $\Rightarrow 3.106 < \pi < 3.215$ | B1 B1 [2] | 3.106, 3.215 |

---
3 Archimedes, about 2200 years ago, used regular polygons inside and outside circles to obtain approximations for $\pi$.
\begin{enumerate}[label=(\roman*)]
\item Fig. 8.1 shows a regular 12 -sided polygon inscribed in a circle of radius 1 unit, centre $\mathrm { O } . \mathrm { AB }$ is one of the sides of the polygon. C is the midpoint of AB . Archimedes used the fact that the circumference of the circle is greater than the perimeter of this polygon.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{c0fcd64b-8ca0-4309-9f58-c23cc4208f4d-2_457_422_457_936}
\captionsetup{labelformat=empty}
\caption{Fig. 8.1}
\end{center}
\end{figure}

(A) Show that $\mathrm { AB } = 2 \sin 15 ^ { \circ }$.\\
(B) Use a double angle formula to express $\cos 30 ^ { \circ }$ in terms of $\sin 15 ^ { \circ }$. Using the exact value of $\cos 30 ^ { \circ }$, show that $\sin 15 ^ { \circ } = \frac { 1 } { 2 } \sqrt { 2 - \sqrt { 3 } }$.\\
(C) Use this result to find an exact expression for the perimeter of the polygon.

Hence show that $\pi > 6 \sqrt { 2 - \sqrt { 3 } }$.
\item In Fig. 8.2, a regular 12-sided polygon lies outside the circle of radius 1 unit, which touches each side of the polygon. F is the midpoint of DE. Archimedes used the fact that the circumference of the circle is less than the perimeter of this polygon.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{c0fcd64b-8ca0-4309-9f58-c23cc4208f4d-2_450_416_1562_940}
\captionsetup{labelformat=empty}
\caption{Fig. 8.2}
\end{center}
\end{figure}

(A) Show that $\mathrm { DE } = 2 \tan 15 ^ { \circ }$.\\
(B) Let $t = \tan 15 ^ { \circ }$. Use a double angle formula to express $\tan 30 ^ { \circ }$ in terms of $t$.

Hence show that $t ^ { 2 } + 2 \sqrt { 3 } t - 1 = 0$.\\
(C) Solve this equation, and hence show that $\pi < 12 ( 2 - \sqrt { 3 } )$.
\item Use the results in parts (i)( $C$ ) and (ii)( $C$ ) to establish upper and lower bounds for the value of $\pi$, giving your answers in decimal form.
\end{enumerate}

\hfill \mbox{\textit{OCR C4  Q3 [19]}}
This paper (7 questions)
View full paper
Q1 6 Q2 7 Q3 19 Q4 7 Q5 6 Q6 7 Q7 6