| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2013 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Trigonometric identities with triangles |
| Difficulty | Moderate -0.3 This is a straightforward multi-step trigonometry problem requiring basic right-angled triangle ratios (tan, cos) and algebraic manipulation. Part (i) involves direct application of trigonometric definitions, while part (ii) requires combining these results—routine for C4 level with no novel insight needed, making it slightly easier than average. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Centre C = (2,3), taxicab radius = 5 (diamond extends to \(x = -0.7\)) | ||
| \(t(P,C) = 5\), so \( | x-2 | + |
| \( | y-3 | = 2.3\), giving \(y = 5.3\) or \(y = 0.7\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Taxicab circle centre \((2,0)\), radius 2: diamond with vertices at \((0,0)\), \((2,2)\), \((4,0)\), \((2,-2)\) drawn correctly | B1 | [1] |
# Question 6(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Centre C = (2,3), taxicab radius = 5 (diamond extends to $x = -0.7$) | | |
| $t(P,C) = 5$, so $|x-2|+|y-3|=5$; at $x=-0.7$: $|-0.7-2|+|y-3|=5 \Rightarrow 2.7+|y-3|=5$ | M1 | Correct substitution |
| $|y-3| = 2.3$, giving $y = 5.3$ or $y = 0.7$ | A1 | Both coordinates needed: $(-0.7, 5.3)$ and $(-0.7, 0.7)$ [2] |
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# Question 6(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Taxicab circle centre $(2,0)$, radius 2: diamond with vertices at $(0,0)$, $(2,2)$, $(4,0)$, $(2,-2)$ drawn correctly | B1 | [1] |6 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]{9bceee25-35bd-448b-a4a2-1a5667be5f11-03_440_524_504_753}
\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 $\mathrm { DE } = 1 + \frac { \tan \phi } { \tan \theta }$.
Section B (36 marks)\\
\hfill \mbox{\textit{OCR MEI C4 2013 Q6 [5]}}