| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Trigonometric identities with triangles |
| Difficulty | Standard +0.3 This is a straightforward multi-step trigonometry problem requiring basic right-angled triangle relationships (SOH CAH TOA) and algebraic manipulation. Part (i) involves simple application of trigonometric ratios, and part (ii) requires substitution and simplification to reach a given result. While it has multiple steps, each step uses standard techniques with no novel insight required, making it slightly easier than average. |
| 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) |
| Answer | Marks | Guidance |
|---|---|---|
| \(AC = \cosec\theta\) | M1 | or \(\frac{1}{\sin\theta}\) |
| \(AD = \cosec\theta \sec\varphi\) | A1 | oe but not if a fraction within a fraction |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(DE = AD\sin(\theta + \varphi)\) | ||
| \(= \cosec\theta \sec\varphi \sin(\theta + \varphi)\) | M1 | \(AD\sin(\theta+\varphi)\) with substitution for their \(AD\) |
| \(= \cosec\theta \sec\varphi(\sin\theta\cos\varphi + \cos\theta\sin\varphi)\) | M1 | correct compound angle formula used |
| \(= \frac{\sin\theta\cos\varphi + \cos\theta\sin\varphi}{\sin\theta\cos\varphi}\) | ||
| \(= 1 + \frac{\cos\theta\sin\varphi}{\sin\theta\cos\varphi}\) | ||
| \(= 1 + \tan\varphi/\tan\theta\) | A1 | simplifying using \(\tan = \sin/\cos\). A0 if no intermediate step, AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 1 + CD\cos\theta\) | M1 | from triangle using \(X\) on \(DE\) where \(CX\) is parallel to \(BE\); \(DX = CD\cos\theta\) and \(CB=1\) |
| \(= 1 + AD\sin\varphi\cos\theta\) | M1 | substituting \(CD = AD\sin\varphi\) and their \(AD\) to reach expression in \(\theta\) and \(\varphi\) only |
| \(= 1 + \cosec\theta\sec\varphi\sin\varphi\cos\theta\) | ||
| \(= 1 + \tan\varphi/\tan\theta\) | A1 | AG |
| [3] |
## Question 2:
### Part (i)
$AC = \cosec\theta$ | M1 | or $\frac{1}{\sin\theta}$
$AD = \cosec\theta \sec\varphi$ | A1 | oe but not if a fraction within a fraction
| [2] |
### Part (ii)
$DE = AD\sin(\theta + \varphi)$ | |
$= \cosec\theta \sec\varphi \sin(\theta + \varphi)$ | M1 | $AD\sin(\theta+\varphi)$ with substitution for their $AD$
$= \cosec\theta \sec\varphi(\sin\theta\cos\varphi + \cos\theta\sin\varphi)$ | M1 | correct compound angle formula used
$= \frac{\sin\theta\cos\varphi + \cos\theta\sin\varphi}{\sin\theta\cos\varphi}$ | |
$= 1 + \frac{\cos\theta\sin\varphi}{\sin\theta\cos\varphi}$ | |
$= 1 + \tan\varphi/\tan\theta$ | A1 | simplifying using $\tan = \sin/\cos$. A0 if no intermediate step, **AG**
**OR** equivalent, e.g. from $DE = CB + CD\cos\theta$:
$= 1 + CD\cos\theta$ | M1 | from triangle using $X$ on $DE$ where $CX$ is parallel to $BE$; $DX = CD\cos\theta$ and $CB=1$
$= 1 + AD\sin\varphi\cos\theta$ | M1 | substituting $CD = AD\sin\varphi$ and their $AD$ to reach expression in $\theta$ and $\varphi$ **only**
$= 1 + \cosec\theta\sec\varphi\sin\varphi\cos\theta$ | |
$= 1 + \tan\varphi/\tan\theta$ | A1 | **AG**
| [3] |
---2 In Fig. 6, $\mathrm { ABC } , \mathrm { ACD }$ and AED are right-angled triangles and $\mathrm { BC } = 1$ unit. Angles CAB and CAD are $\theta$ and $\phi$ respectively.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c8ea5913-c527-40e7-bfcc-c1c2df544e04-2_452_535_437_781}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}
(i) Find AC and AD in terms of $\theta$ and $\phi$.\\
(ii) Hence show that $\mathrm { DE } = 1 + \frac { \tan \phi } { \tan \theta }$.
\hfill \mbox{\textit{OCR MEI C4 Q2 [5]}}