| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Sequential triangle calculations (basic) |
| Difficulty | Easy -1.2 This is a straightforward two-part question requiring direct application of the sine rule to find a missing side, followed by standard area formula (1/2)ab sin C. Both are routine C2-level calculations with no problem-solving or conceptual challenges—simpler than average A-level questions. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C) |
| Answer | Marks |
|---|---|
| \(\angle BAC = 180 - (107 + 31) = 42\) | B1 |
| \(\frac{BC}{\sin 42} = \frac{12.6}{\sin 31}\) | M1 |
| \(BC = \frac{12.6 \sin 42}{\sin 31} = 16.4 \text{ cm (3sf)}\) | A1 |
| Area \(= \frac{1}{2} \times 12.6 \times 16.37 \times \sin 107 = 98.6 \text{ cm}^2 \text{ (3sf)}\) | M1 A1 |
$\angle BAC = 180 - (107 + 31) = 42$ | B1 |
$\frac{BC}{\sin 42} = \frac{12.6}{\sin 31}$ | M1 |
$BC = \frac{12.6 \sin 42}{\sin 31} = 16.4 \text{ cm (3sf)}$ | A1 |
Area $= \frac{1}{2} \times 12.6 \times 16.37 \times \sin 107 = 98.6 \text{ cm}^2 \text{ (3sf)}$ | M1 A1 |1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{089f5506-94ac-489f-b219-e67fa6ca834f-2_383_707_246_488}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows triangle $A B C$ in which $A B = 12.6 \mathrm {~cm} , \angle A B C = 107 ^ { \circ }$ and $\angle A C B = 31 ^ { \circ }$.\\
Find, to 3 significant figures,
\begin{enumerate}[label=(\alph*)]
\item the length $B C$,
\item the area of triangle $A B C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q1 [5]}}