| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2010 |
| Session | January |
| Marks | 2 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Applied context with trigonometry |
| Difficulty | Standard +0.3 This is a structured multi-part question testing standard C4 content (addition formulae, double angle identities, and R-formula). Parts (i)-(ii) involve routine geometry and formula manipulation with clear guidance ('show that'). Part (iii) requires setting dh/dθ=0 or recognizing horizontal OB condition. Part (iv) is a textbook R-formula application. While it requires multiple techniques, each step follows standard procedures with scaffolding, making it slightly easier than average. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.05b Sine and cosine rules: including ambiguous case1.05l Double angle formulae: and compound angle formulae1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Longer strings to analyse so letter frequency more transparent. Or there are fewer 2-letter keywords to check | B2 |
### Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Longer strings to analyse so letter frequency more transparent. Or there are fewer 2-letter keywords to check | B2 | |6 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}
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\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 2010 Q6 [2]}}