| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Addition & Double Angle Formulae |
| Type | Applied context with trigonometry |
| Difficulty | Standard +0.3 This is a standard applied trigonometry question requiring the tangent addition formula, implicit differentiation, and optimization. While it involves multiple steps (3 parts, likely 8-10 marks total), each technique is routine for C4: tan(β-α) formula application, implicit differentiation of tan θ, and finding maxima by setting derivative to zero. The geometry setup is straightforward, and the algebraic manipulation, though somewhat lengthy, follows standard patterns without requiring novel insight. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.05q Trig in context: vectors, kinematics, forces1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(A\hat{O}P = 180 - \beta = 180 - \alpha - \theta\) | M1 | Use of sum of angles in triangle OPT and AOP |
| \(\Rightarrow \beta = \alpha + \theta\) | M1 | |
| \(\Rightarrow \theta = \beta - \alpha\) | E1 | SC B1 for \(\beta = \alpha + \theta\), \(\theta = \beta - \alpha\) no justification |
| \(\tan\theta = \tan(\beta - \alpha) = \dfrac{\tan\beta - \tan\alpha}{1 + \tan\beta\tan\alpha}\) | M1 | Use of Compound angle formula |
| \(= \dfrac{\frac{y}{10} - \frac{y}{16}}{1 + \frac{y}{10}\cdot\frac{y}{16}}\) | A1 | Substituting values for \(\tan\alpha\) and \(\tan\beta\) |
| \(= \dfrac{16y - 10y}{160 + y^2} = \dfrac{6y}{160 + y^2}\) * | E1 | |
| When \(y = 6\), \(\tan\theta = 36/196\) | M1 | www |
| \(\Rightarrow \theta = 10.4°\) | A1 cao [8] | accept radians |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sec^2\theta\dfrac{d\theta}{dy} = \dfrac{(160 + y^2)6 - 6y \cdot 2y}{(160 + y^2)^2}\) | M1 | \(\sec^2\theta\dfrac{d\theta}{dy} = \ldots\) |
| \(= \dfrac{6(160 + y^2 - 2y^2)}{(160 + y^2)^2}\) | M1 | quotient rule |
| A1 | correct expression | |
| \(\Rightarrow \dfrac{d\theta}{dy} = \dfrac{6(160 - y^2)}{(160 + y^2)^2}\cos^2\theta\) * | A1 | simplifying numerator www |
| E1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(d\theta/dy = 0\) when \(160 - y^2 = 0\) | M1 | |
| \(\Rightarrow y^2 = 160\) | A1 | oe |
| \(\Rightarrow y = 12.65\) | ||
| When \(y = 12.65\), \(\tan\theta = 0.237\ldots\) | M1 | |
| \(\Rightarrow \theta = 13.3°\) | A1cao [4] | accept radians |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $A\hat{O}P = 180 - \beta = 180 - \alpha - \theta$ | M1 | Use of sum of angles in triangle OPT and AOP |
| $\Rightarrow \beta = \alpha + \theta$ | M1 | |
| $\Rightarrow \theta = \beta - \alpha$ | E1 | SC B1 for $\beta = \alpha + \theta$, $\theta = \beta - \alpha$ no justification |
| $\tan\theta = \tan(\beta - \alpha) = \dfrac{\tan\beta - \tan\alpha}{1 + \tan\beta\tan\alpha}$ | M1 | Use of Compound angle formula |
| $= \dfrac{\frac{y}{10} - \frac{y}{16}}{1 + \frac{y}{10}\cdot\frac{y}{16}}$ | A1 | Substituting values for $\tan\alpha$ and $\tan\beta$ |
| $= \dfrac{16y - 10y}{160 + y^2} = \dfrac{6y}{160 + y^2}$ * | E1 | |
| When $y = 6$, $\tan\theta = 36/196$ | M1 | www |
| $\Rightarrow \theta = 10.4°$ | A1 cao [8] | accept radians |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sec^2\theta\dfrac{d\theta}{dy} = \dfrac{(160 + y^2)6 - 6y \cdot 2y}{(160 + y^2)^2}$ | M1 | $\sec^2\theta\dfrac{d\theta}{dy} = \ldots$ |
| $= \dfrac{6(160 + y^2 - 2y^2)}{(160 + y^2)^2}$ | M1 | quotient rule |
| | A1 | correct expression |
| $\Rightarrow \dfrac{d\theta}{dy} = \dfrac{6(160 - y^2)}{(160 + y^2)^2}\cos^2\theta$ * | A1 | simplifying numerator www |
| | E1 [5] | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $d\theta/dy = 0$ when $160 - y^2 = 0$ | M1 | |
| $\Rightarrow y^2 = 160$ | A1 | oe |
| $\Rightarrow y = 12.65$ | | |
| When $y = 12.65$, $\tan\theta = 0.237\ldots$ | M1 | |
| $\Rightarrow \theta = 13.3°$ | A1cao [4] | accept radians |
---2 In a game of rugby, a kick is to be taken from a point P (see Fig. 7). P is a perpendicular distance $y$ metres from the line TOA. Other distances and angles are as shown.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{7a52b6ce-a0cc-421d-8eae-3b6cf098e381-2_511_630_449_750}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{center}
\end{figure}
(i) Show that $\theta = \beta - \alpha$, and hence that $\tan \theta = \frac { 6 y } { 160 + y ^ { 2 } }$.
Calculate the angle $\theta$ when $y = 6$.\\
(ii) By differentiating implicitly, show that $\frac { \mathrm { d } \theta } { \mathrm { d } y } = \frac { 6 \left( 160 - y ^ { 2 } \right) } { \left( 160 + y ^ { 2 } \right) ^ { 2 } } \cos ^ { 2 } \theta$.\\
(iii) Use this result to find the value of $y$ that maximises the angle $\theta$. Calculate this maximum value of $\theta$. [You need not verify that this value is indeed a maximum.]
\hfill \mbox{\textit{OCR MEI C4 Q2 [17]}}