| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Quadrilateral with diagonal |
| Difficulty | Standard +0.3 This is a straightforward application of the cosine rule to find an angle in triangle ABC, followed by using Pythagoras identity for sine, then applying the sine rule in triangle ACD using the parallel line property. All steps are standard techniques with clear signposting, making it slightly easier than average despite being multi-part. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\cos BCA = \frac{5^2 + 6^2 - 9^2}{2 \times 5 \times 6} = -\frac{1}{4}\) | M1, M1, A1, B1 | For relevant use of the correct cosine formula; For attempt to rearrange formula; For obtaining the given value correctly; For correct answer for \(\sin BCA\) in any form OR |
| \(\sin BCA = \frac{3}{4}\sqrt{2} = 0.9428 \ldots\) | M1, M1, A1, B1 | For substituting \(\cos BCA = -\frac{\sqrt{3}}{4}\); For attempt at evaluation; For full verification; For correct answer for \(\sin BCA\) in any form |
| (ii) Angles \(BCA\) and \(CAD\) are equal | B1 | For stating, using or implying the equal angles |
| So \(\sin ADC = \frac{1}{4}\sin CAD = \frac{1}{x} \times \frac{\sqrt{8}}{4} = \frac{1}{x} \times \sqrt{8} = \frac{1}{\sqrt{2}}\) | M1, A1, A1 | For correct use of the sine rule in \(\triangle ADC\) (sides must be numerical, angles may still be in letters); For a correct equation from their value in (i); For correct answer, from correct working |
| \(\Rightarrow ADC = 18.3°\) | A1 |
**(i)** $\cos BCA = \frac{5^2 + 6^2 - 9^2}{2 \times 5 \times 6} = -\frac{1}{4}$ | M1, M1, A1, B1 | For relevant use of the correct cosine formula; For attempt to rearrange formula; For obtaining the given value correctly; For correct answer for $\sin BCA$ in any form OR
$\sin BCA = \frac{3}{4}\sqrt{2} = 0.9428 \ldots$ | M1, M1, A1, B1 | For substituting $\cos BCA = -\frac{\sqrt{3}}{4}$; For attempt at evaluation; For full verification; For correct answer for $\sin BCA$ in any form
**(ii)** Angles $BCA$ and $CAD$ are equal | B1 | For stating, using or implying the equal angles
So $\sin ADC = \frac{1}{4}\sin CAD = \frac{1}{x} \times \frac{\sqrt{8}}{4} = \frac{1}{x} \times \sqrt{8} = \frac{1}{\sqrt{2}}$ | M1, A1, A1 | For correct use of the sine rule in $\triangle ADC$ (sides must be numerical, angles may still be in letters); For a correct equation from their value in (i); For correct answer, from correct working
$\Rightarrow ADC = 18.3°$ | A1 |
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\includegraphics[max width=\textwidth, alt={}, center]{387a37c4-0997-484c-8e28-954639169ebe-3_309_1084_269_532}
In the diagram, $A B C D$ is a quadrilateral in which $A D$ is parallel to $B C$. It is given that $A B = 9 , B C = 6$, $C A = 5$ and $C D = 15$.\\
(i) Show that $\cos B C A = - \frac { 1 } { 3 }$, and hence find the value of $\sin B C A$.\\
(ii) Find the angle $A D C$ correct to the nearest $0.1 ^ { \circ }$.
\hfill \mbox{\textit{OCR C2 2005 Q4 [8]}}