| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Trigonometric identities with triangles |
| Difficulty | Challenging +1.2 This is a structured multi-part question with clear scaffolding through parts (i)-(v). While it involves 3D geometry visualization and trigonometric manipulation, each step is guided with results to show. The trigonometry is standard (tan addition formula, sec identities) and the algebraic manipulation is routine. The 3D setup requires some spatial reasoning but the mathematical techniques are all Core 4 standard material with no novel insights required. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C)1.05l Double angle formulae: and compound angle formulae1.05q Trig in context: vectors, kinematics, forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(AC = 5\sec\alpha\) | B1 | oe |
| \(CF = AC\tan\beta = 5\sec\alpha\tan\beta\) | M1, E1 | \(AC\tan\beta\) |
| \(GF = 2CF = 10\sec\alpha\tan\beta\) | ||
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(CE = BE - BC = 5\tan(\alpha+\beta) - 5\tan\alpha\) | E1 | |
| \(= 5\left(\dfrac{\tan\alpha + \tan\beta}{1-\tan\alpha\tan\beta} - \tan\alpha\right)\) | M1 | Compound angle formula |
| \(= 5\left(\dfrac{\tan\alpha + \tan\beta - \tan\alpha + \tan^2\alpha\tan\beta}{1-\tan\alpha\tan\beta}\right)\) | M1 | Combining fractions |
| \(= \dfrac{5(1+\tan^2\alpha)\tan\beta}{1-\tan\alpha\tan\beta}\) | DM1 | \(\sec^2 = 1 + \tan^2\) |
| \(= \dfrac{5\tan\beta\sec^2\alpha}{1-\tan\alpha\tan\beta}\) | E1 | |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\sec^2 45° = 2,\quad \tan 45° = 1\) | B1 | Used |
| \(CE = \dfrac{5t \times 2}{1-t} = \dfrac{10t}{1-t}\) | M1, A1 | Substitution for both in CE or CD oe; for both |
| \(CD = \dfrac{10t}{1+t}\) | ||
| \(DE = \dfrac{10t}{1-t} + \dfrac{10t}{1+t} = 10t\left(\dfrac{1}{1-t}+\dfrac{1}{1+t}\right)\) | M1 | Adding their CE and CD |
| \(= 10t\left(\dfrac{1+t+1-t}{(1-t)(1+t)}\right) = \dfrac{20t}{1-t^2}\) | E1 | |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\cos 45° = \dfrac{1}{\sqrt{2}} \Rightarrow \sec\alpha = \sqrt{2}\) | M1 | |
| \(GF = 10\sqrt{2}\tan\beta = 10\sqrt{2}\,t\) | E1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(DE = 2GF \Rightarrow \dfrac{20t}{1-t^2} = 20\sqrt{2}\,t\) | E1 | |
| \(\Rightarrow 1-t^2 = \dfrac{1}{\sqrt{2}} \Rightarrow t^2 = 1 - \dfrac{1}{\sqrt{2}}\) | M1 | invtan \(t\) |
| \(\Rightarrow t = 0.541 \Rightarrow \beta = 28.4°\) | A1 | |
| [3] |
## Question 3:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $AC = 5\sec\alpha$ | B1 | oe |
| $CF = AC\tan\beta = 5\sec\alpha\tan\beta$ | M1, E1 | $AC\tan\beta$ |
| $GF = 2CF = 10\sec\alpha\tan\beta$ | | |
| **[3]** | | |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $CE = BE - BC = 5\tan(\alpha+\beta) - 5\tan\alpha$ | E1 | |
| $= 5\left(\dfrac{\tan\alpha + \tan\beta}{1-\tan\alpha\tan\beta} - \tan\alpha\right)$ | M1 | Compound angle formula |
| $= 5\left(\dfrac{\tan\alpha + \tan\beta - \tan\alpha + \tan^2\alpha\tan\beta}{1-\tan\alpha\tan\beta}\right)$ | M1 | Combining fractions |
| $= \dfrac{5(1+\tan^2\alpha)\tan\beta}{1-\tan\alpha\tan\beta}$ | DM1 | $\sec^2 = 1 + \tan^2$ |
| $= \dfrac{5\tan\beta\sec^2\alpha}{1-\tan\alpha\tan\beta}$ | E1 | |
| **[5]** | | |
### Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sec^2 45° = 2,\quad \tan 45° = 1$ | B1 | Used |
| $CE = \dfrac{5t \times 2}{1-t} = \dfrac{10t}{1-t}$ | M1, A1 | Substitution for both in CE or CD oe; for both |
| $CD = \dfrac{10t}{1+t}$ | | |
| $DE = \dfrac{10t}{1-t} + \dfrac{10t}{1+t} = 10t\left(\dfrac{1}{1-t}+\dfrac{1}{1+t}\right)$ | M1 | Adding their CE and CD |
| $= 10t\left(\dfrac{1+t+1-t}{(1-t)(1+t)}\right) = \dfrac{20t}{1-t^2}$ | E1 | |
| **[5]** | | |
### Part (iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\cos 45° = \dfrac{1}{\sqrt{2}} \Rightarrow \sec\alpha = \sqrt{2}$ | M1 | |
| $GF = 10\sqrt{2}\tan\beta = 10\sqrt{2}\,t$ | E1 | |
| **[2]** | | |
### Part (v):
| Answer | Mark | Guidance |
|--------|------|----------|
| $DE = 2GF \Rightarrow \dfrac{20t}{1-t^2} = 20\sqrt{2}\,t$ | E1 | |
| $\Rightarrow 1-t^2 = \dfrac{1}{\sqrt{2}} \Rightarrow t^2 = 1 - \dfrac{1}{\sqrt{2}}$ | M1 | invtan $t$ |
| $\Rightarrow t = 0.541 \Rightarrow \beta = 28.4°$ | A1 | |
| **[3]** | | |
---3 Fig. 8 shows a searchlight, mounted at a point A, 5 metres above level ground. Its beam is in the shape of a cone with axis AC , where C is on the ground. AC is angled at $\alpha$ to the vertical. The beam produces an oval-shaped area of light on the ground, of length DE . The width of the oval at C is GF . Angles DAC, EAC, FAC and GAC are all $\beta$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f5ab2a9d-3366-4b9b-86b0-85d5d923953c-2_695_877_503_242}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f5ab2a9d-3366-4b9b-86b0-85d5d923953c-2_333_799_505_1048}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
In the following, all lengths are in metres.\\
(i) Find AC in terms of $\alpha$, and hence show that $\mathrm { GF } = 10 \sec \alpha \tan \beta$.\\
(ii) Show that $\mathrm { CE } = 5 ( \tan ( \alpha + \beta ) - \tan \alpha )$.
$$\text { Hence show that } \mathrm { CE } = \frac { 5 \tan \beta \sec ^ { 2 } \alpha } { 1 - \tan \alpha \tan \beta } \text {. }$$
Similarly, it can be shown that $\mathrm { CD } = \frac { 5 \tan \beta \sec ^ { 2 } \alpha } { 1 + \tan \alpha \tan \beta }$. [You are not required to derive this result.]\\
You are now given that $\alpha = 45 ^ { \circ }$ and that $\tan \beta = t$.\\
(iii) Find CE and CD in terms of $t$. Hence show that $\mathrm { DE } = \frac { 20 t } { 1 - t ^ { 2 } }$.\\
(iv) Show that $\mathrm { GF } = 10 \sqrt { 2 } t$.
For a certain value of $\beta , \mathrm { DE } = 2 \mathrm { GF }$.\\
(v) Show that $t ^ { 2 } = 1 - \frac { 1 } { \sqrt { 2 } }$.
Hence find this value of $\beta$.
\hfill \mbox{\textit{OCR MEI C4 Q3 [18]}}