| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2008 |
| Session | January |
| Marks | 6 |
| 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 (part a) and triangle area formula (part b). Both are standard textbook exercises with no problem-solving required—students simply substitute given values into memorized formulas. The 'show that' in part (a) provides the answer, making it even more routine. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{BC}{\sin 72} = \frac{18.7}{\sin 50}\) \([=24.4\ldots]\) | M1 | Use of the sine rule |
| \(BC = \frac{18.7\sin 72}{\sin 50}\) | m1 | Rearrangement |
| \((BC)=23.21(6\ldots)\{= 23.2\) to nearest \(0.1\)cm\(\}\) | A1 | AG Need \(>1\)dp if using cm eg 23.21 or 23.22; at least 1dp if using mm — Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Angle \(C = 180° - (50°+72°) = 58°\) | M1 | Valid method to find either angle \(C\) (PI eg by \(\sin C = 0.848(04\ldots)\)) or side \(AB\) |
| Area of triangle \(= 0.5\times18.7\times23.2\ldots\times\sin C\) | M1 | OE eg \(0.5\times18.7\times AB\times\sin72°\) |
| \(\ldots = 184\) cm² | A1 | Accept 183.8 to 184.2. Condone missing/wrong units — Total: 3 |
## Question 3:
### Part (a)
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{BC}{\sin 72} = \frac{18.7}{\sin 50}$ $[=24.4\ldots]$ | M1 | Use of the sine rule |
| $BC = \frac{18.7\sin 72}{\sin 50}$ | m1 | Rearrangement |
| $(BC)=23.21(6\ldots)\{= 23.2$ to nearest $0.1$cm$\}$ | A1 | AG Need $>1$dp if using cm eg 23.21 or 23.22; at least 1dp if using mm — **Total: 3** |
### Part (b)
| Working | Mark | Guidance |
|---------|------|----------|
| Angle $C = 180° - (50°+72°) = 58°$ | M1 | Valid method to find either angle $C$ (PI eg by $\sin C = 0.848(04\ldots)$) or side $AB$ |
| Area of triangle $= 0.5\times18.7\times23.2\ldots\times\sin C$ | M1 | OE eg $0.5\times18.7\times AB\times\sin72°$ |
| $\ldots = 184$ cm² | A1 | Accept 183.8 to 184.2. Condone missing/wrong units — **Total: 3** |
---3 The diagram shows a triangle $A B C$. The length of $A C$ is 18.7 cm , and the sizes of angles $B A C$ and $A B C$ are $72 ^ { \circ }$ and $50 ^ { \circ }$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Show that the length of $B C = 23.2 \mathrm {~cm}$, correct to the nearest 0.1 cm .
\item Calculate the area of triangle $A B C$, giving your answer to the nearest $\mathrm { cm } ^ { 2 }$.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2008 Q3 [6]}}