OCR MEI C2 2012 June — Question 3 5 marks

Exam BoardOCR MEI
ModuleC2 (Core Mathematics 2)
Year2012
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSine and Cosine Rules
TypePoint on side of triangle
DifficultyModerate -0.3 This is a straightforward application of the cosine rule (part i) and standard triangle area formula (part ii). The key insight that angle ACB = 180° - 53.4° is immediate from BCD being a straight line. Both parts are direct one-step calculations with no problem-solving required, making this slightly easier than average but not trivial since it involves applying the cosine rule rather than just basic trigonometry.
Spec1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C)

3 \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{8f7413d8-2814-4d5c-bec0-ce118fec80eb-2_592_693_845_502} \captionsetup{labelformat=empty} \caption{Fig. 3}
\end{figure} \section*{Not to scale} In Fig. 3, BCD is a straight line. \(\mathrm { AC } = 9.8 \mathrm {~cm} , \mathrm { BC } = 7.3 \mathrm {~cm}\) and \(\mathrm { CD } = 6.4 \mathrm {~cm}\); angle \(\mathrm { ACD } = 53.4 ^ { \circ }\).
  1. Calculate the length AD .
  2. Calculate the area of triangle ABC .

3

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{8f7413d8-2814-4d5c-bec0-ce118fec80eb-2_592_693_845_502}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}

\section*{Not to scale}
In Fig. 3, BCD is a straight line. $\mathrm { AC } = 9.8 \mathrm {~cm} , \mathrm { BC } = 7.3 \mathrm {~cm}$ and $\mathrm { CD } = 6.4 \mathrm {~cm}$; angle $\mathrm { ACD } = 53.4 ^ { \circ }$.\\
(i) Calculate the length AD .\\
(ii) Calculate the area of triangle ABC .

\hfill \mbox{\textit{OCR MEI C2 2012 Q3 [5]}}