OCR MEI C4 — Question 1 16 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Marks16
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAddition & Double Angle Formulae
TypeApplied context with trigonometry
DifficultyStandard +0.3 This is a structured multi-part question testing standard C4 content (addition formulae, double angle identities, and R-formula). Each part guides students through the solution with clear scaffolding. The geometry requires careful angle-chasing but is straightforward, and the trigonometric manipulations are routine textbook exercises. Slightly easier than average due to the extensive scaffolding.
Spec1.05b 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 theta1.05l Double angle formulae: and compound angle formulae1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc
1 In Fig. 6, OAB is a thin bent rod, with \(\mathrm { OA } = a\) metres, \(\mathrm { AB } = b\) metres and angle \(\mathrm { OAB } = 120 ^ { \circ }\). The bent rod lies in a vertical plane. OA makes an angle \(\theta\) above the horizontal. The vertical height BD of B above O is \(h\) metres. The horizontal through A meets BD at C and the vertical through A meets OD at E . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{9ac55ae6-7a7f-47d0-a363-92da179be4ca-1_427_898_464_683} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure}
  1. Find angle BAC in terms of \(\theta\). Hence show that $$h = a \sin \theta + b \sin \left( \theta - 60 ^ { \circ } \right) .$$
  2. Hence show that \(h = \left( a + \frac { 1 } { 2 } b \right) \sin \theta - \frac { \sqrt { 3 } } { 2 } b \cos \theta\). The rod now rotates about O , so that \(\theta\) varies. You may assume that the formulae for \(h\) in parts (i) and (ii) remain valid.
  3. Show that OB is horizontal when \(\tan \theta = \frac { \sqrt { 3 } b } { 2 a + b }\). In the case when \(a = 1\) and \(b = 2 , h = 2 \sin \theta - \sqrt { 3 } \cos \theta\).
  4. Express \(2 \sin \theta - \sqrt { 3 } \cos \theta\) in the form \(R \sin ( \theta - \alpha )\). Hence, for this case, write down the maximum value of \(h\) and the corresponding value of \(\theta\).
Question 1
(i)
AnswerMarks
\(\text{BAC} = 120 - 90 - (90 - \theta) = \theta - 60\)B1
\(\text{BC} = b \sin(\theta - 60)\)M1
\(\text{CD} = \text{AE} = a \sin \theta\)E1
\(h = \text{BC} + \text{CD} = a \sin \theta + b \sin(\theta - 60°)\)[3]
(ii)
AnswerMarks
\(h = a \sin \theta + b \sin(\theta - 60°)\)M1
correct compound angle formula
AnswerMarks
\(= a \sin \theta + b(\sin\theta \cos 60 - \cos\theta \sin 60)\)M1
\(\sin 60 = \frac{\sqrt{3}}{2}\), \(\cos 60 = \frac{1}{2}\) used
AnswerMarks
\(= a \sin \theta + \frac{1}{2}b \sin \theta - \frac{\sqrt{3}}{2}b \cos \theta\)E1
\(= \left(a + \frac{1}{2}b\right) \sin\theta - \frac{\sqrt{3}}{2}b \cos\theta\)[3]
(iii)
AnswerMarks
OB horizontal when \(h = 0\)M1
\(\left(a + \frac{1}{2}b\right) \sin\theta - \frac{\sqrt{3}}{2}b \cos\theta = 0\)M1
\(\left(a + \frac{1}{2}b\right) \sin\theta = \frac{\sqrt{3}}{2}b \cos\theta\)
\(\frac{\sin\theta}{\cos\theta} = \tan\theta\)
\(\tan\theta = \frac{\frac{\sqrt{3}}{2}b}{a + \frac{1}{2}b} = \frac{\sqrt{3}b}{2a + b}\)E1
[3]
(iv)
AnswerMarks
\(2\sin\theta - \sqrt{3}\cos\theta = R\sin(\theta - \alpha)\)M1
\(= R(\sin\theta \cos\alpha - \cos\theta \sin\alpha)\)
\(R \cos \alpha = 2\), \(R \sin \alpha = \sqrt{3}\)B1
\(R^2 = 2^2 + (\sqrt{3})^2 = 7\), \(R = \sqrt{7} = 2.646\) mM1A1
\(\tan \alpha = \frac{\sqrt{3}}{2}\), \(\alpha = 40.9°\)B1ft
So \(h = \sqrt{7} \sin(\theta - 40.9°)\)
\(h_{\max} = \sqrt{7} = 2.646\) m when \(\theta - 40.9° = 90°\)M1
\(\theta = 130.9°\)A1
[7]
AnswerMarks
\(2\sin\theta - \sqrt{3}\cos\theta = R\sin(\theta - \alpha)\)M1
\(= R(\sin\theta \cos\alpha - \cos\theta \sin\alpha)\)
\(3\cos\theta + 4\sin\theta = R\cos(\theta - \alpha)\)
\(= R(\cos\theta \cos\alpha + \sin\theta \sin\alpha)\)
\(\Rightarrow R \cos \alpha = 3\), \(R \sin \alpha = 4\)B1
\(\Rightarrow R^2 = 3^2 + 4^2 = 25 \Rightarrow R = 5\)M1A1
\(\tan \alpha = \frac{4}{3} \Rightarrow \alpha = 0.9273\)
\(5\cos(\theta - 0.9273) = 2\)M1
\(\cos(\theta - 0.9273) = \frac{2}{5}\)
\(\theta - 0.9273 = 1.1593, -1.1593\)A1
\(\theta = 2.087, -0.232\)A1
\(R = 5\) cwo and no others in the range[7] 
# Question 1

## (i)
$\text{BAC} = 120 - 90 - (90 - \theta) = \theta - 60$ | B1

$\text{BC} = b \sin(\theta - 60)$ | M1

$\text{CD} = \text{AE} = a \sin \theta$ | E1

$h = \text{BC} + \text{CD} = a \sin \theta + b \sin(\theta - 60°)$ | [3]

## (ii)
$h = a \sin \theta + b \sin(\theta - 60°)$ | M1
correct compound angle formula

$= a \sin \theta + b(\sin\theta \cos 60 - \cos\theta \sin 60)$ | M1
$\sin 60 = \frac{\sqrt{3}}{2}$, $\cos 60 = \frac{1}{2}$ used

$= a \sin \theta + \frac{1}{2}b \sin \theta - \frac{\sqrt{3}}{2}b \cos \theta$ | E1

$= \left(a + \frac{1}{2}b\right) \sin\theta - \frac{\sqrt{3}}{2}b \cos\theta$ | [3]

## (iii)
OB horizontal when $h = 0$ | M1

$\left(a + \frac{1}{2}b\right) \sin\theta - \frac{\sqrt{3}}{2}b \cos\theta = 0$ | M1

$\left(a + \frac{1}{2}b\right) \sin\theta = \frac{\sqrt{3}}{2}b \cos\theta$ | 

$\frac{\sin\theta}{\cos\theta} = \tan\theta$ |

$\tan\theta = \frac{\frac{\sqrt{3}}{2}b}{a + \frac{1}{2}b} = \frac{\sqrt{3}b}{2a + b}$ | E1
[3]

## (iv)
$2\sin\theta - \sqrt{3}\cos\theta = R\sin(\theta - \alpha)$ | M1

$= R(\sin\theta \cos\alpha - \cos\theta \sin\alpha)$ | 

$R \cos \alpha = 2$, $R \sin \alpha = \sqrt{3}$ | B1

$R^2 = 2^2 + (\sqrt{3})^2 = 7$, $R = \sqrt{7} = 2.646$ m | M1A1

$\tan \alpha = \frac{\sqrt{3}}{2}$, $\alpha = 40.9°$ | B1ft

So $h = \sqrt{7} \sin(\theta - 40.9°)$ | 

$h_{\max} = \sqrt{7} = 2.646$ m when $\theta - 40.9° = 90°$ | M1

$\theta = 130.9°$ | A1
[7]

---

$2\sin\theta - \sqrt{3}\cos\theta = R\sin(\theta - \alpha)$ | M1

$= R(\sin\theta \cos\alpha - \cos\theta \sin\alpha)$ | 

$3\cos\theta + 4\sin\theta = R\cos(\theta - \alpha)$ | 

$= R(\cos\theta \cos\alpha + \sin\theta \sin\alpha)$ | 

$\Rightarrow R \cos \alpha = 3$, $R \sin \alpha = 4$ | B1

$\Rightarrow R^2 = 3^2 + 4^2 = 25 \Rightarrow R = 5$ | M1A1

$\tan \alpha = \frac{4}{3} \Rightarrow \alpha = 0.9273$ | 

$5\cos(\theta - 0.9273) = 2$ | M1

$\cos(\theta - 0.9273) = \frac{2}{5}$ | 

$\theta - 0.9273 = 1.1593, -1.1593$ | A1

$\theta = 2.087, -0.232$ | A1
$R = 5$ cwo and no others in the range | [7]
1 In Fig. 6, OAB is a thin bent rod, with $\mathrm { OA } = a$ metres, $\mathrm { AB } = b$ metres and angle $\mathrm { OAB } = 120 ^ { \circ }$. The bent rod lies in a vertical plane. OA makes an angle $\theta$ above the horizontal. The vertical height BD of B above O is $h$ metres. The horizontal through A meets BD at C and the vertical through A meets OD at E .

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{9ac55ae6-7a7f-47d0-a363-92da179be4ca-1_427_898_464_683}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}

(i) Find angle BAC in terms of $\theta$. Hence show that

$$h = a \sin \theta + b \sin \left( \theta - 60 ^ { \circ } \right) .$$

(ii) Hence show that $h = \left( a + \frac { 1 } { 2 } b \right) \sin \theta - \frac { \sqrt { 3 } } { 2 } b \cos \theta$.

The rod now rotates about O , so that $\theta$ varies. You may assume that the formulae for $h$ in parts (i) and (ii) remain valid.\\
(iii) Show that OB is horizontal when $\tan \theta = \frac { \sqrt { 3 } b } { 2 a + b }$.

In the case when $a = 1$ and $b = 2 , h = 2 \sin \theta - \sqrt { 3 } \cos \theta$.\\
(iv) Express $2 \sin \theta - \sqrt { 3 } \cos \theta$ in the form $R \sin ( \theta - \alpha )$. Hence, for this case, write down the maximum value of $h$ and the corresponding value of $\theta$.

\hfill \mbox{\textit{OCR MEI C4  Q1 [16]}}
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