| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Point on side of triangle |
| Difficulty | Moderate -0.3 Part (a) is a straightforward sine rule application with angles and one side given, requiring minimal problem-solving. Part (b) requires finding the midpoint then using cosine rule, which adds a modest step but remains a standard two-part question testing routine techniques without novel insight. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Angle \(ACB = 180 - 72 - 48 = 60°\) | B1 | Correct angle found |
| \(\frac{AC}{\sin 48°} = \frac{20}{\sin 60°}\) | M1 | Correct sine rule application |
| \(AC = \frac{20 \sin 48°}{\sin 60°} = \frac{20 \times 0.7431...}{0.8660...} = 15.6\) cm | A1 | Correct answer shown to 3 s.f. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(BM = 10\) cm (M is midpoint of BC) | B1 | Correct value stated |
| \(AM^2 = AB^2 + BM^2 - 2(AB)(BM)\cos 48°\) | M1 | Correct cosine rule applied in triangle \(ABM\) |
| \(AB = \frac{20\sin72°}{\sin60°} = 21.94...\) cm | A1 | Correct value of \(AB\) |
| \(AM^2 = (21.94)^2 + 100 - 2(21.94)(10)\cos48°\) | ||
| \(AM = 16.4\) cm | A1 | Correct final answer to 3 s.f. |
# Question 2:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Angle $ACB = 180 - 72 - 48 = 60°$ | B1 | Correct angle found |
| $\frac{AC}{\sin 48°} = \frac{20}{\sin 60°}$ | M1 | Correct sine rule application |
| $AC = \frac{20 \sin 48°}{\sin 60°} = \frac{20 \times 0.7431...}{0.8660...} = 15.6$ cm | A1 | Correct answer shown to 3 s.f. |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $BM = 10$ cm (M is midpoint of BC) | B1 | Correct value stated |
| $AM^2 = AB^2 + BM^2 - 2(AB)(BM)\cos 48°$ | M1 | Correct cosine rule applied in triangle $ABM$ |
| $AB = \frac{20\sin72°}{\sin60°} = 21.94...$ cm | A1 | Correct value of $AB$ |
| $AM^2 = (21.94)^2 + 100 - 2(21.94)(10)\cos48°$ | | |
| $AM = 16.4$ cm | A1 | Correct final answer to 3 s.f. |
---2 The diagram shows a triangle $A B C$.
The size of angle $B A C$ is $72 ^ { \circ }$ and the size of angle $A B C$ is $48 ^ { \circ }$. The length of $B C$ is 20 cm .
\begin{enumerate}[label=(\alph*)]
\item Show that the length of $A C$ is 15.6 cm , correct to three significant figures.
\item The midpoint of $B C$ is $M$. Calculate the length of $A M$, giving your answer, in cm , to three significant figures.\\[0pt]
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2015 Q2 [7]}}