| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Triangle with trigonometric identities |
| Difficulty | Moderate -0.8 This is a straightforward application of standard formulas: (a) uses area = ½ab sin C directly with given values, (b) requires the Pythagorean identity sin²θ + cos²θ = 1 with a simple exact value, and (c) applies the cosine rule. All three parts are routine textbook exercises with no problem-solving or insight required, making it easier than average but not trivial since it requires multiple techniques. |
| Spec | 1.05c Area of triangle: using 1/2 ab sin(C)1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Area} = \frac{1}{2}(26)(31.5)\sin\theta = \frac{1}{2}(26)(31.5)\times\frac{5}{13} = 157.5 \text{ cm}^2\) | M1, A1 | \(\frac{1}{2}(26)(31.5)\sin(\theta)\) stated or used; OE e.g. \(\frac{315}{2}\); Condone AWRT 157.50; NMS: 157.5 or AWRT 157.50 scores B2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos\theta = \frac{12}{13}\) | B1 | \(\frac{12}{13}\) OE exact fraction |
| Answer | Marks | Guidance |
|---|---|---|
| \(AC = \sqrt{156.25} = 12.5\text{ cm}\) | M1, m1, A1 | RHS of cosine rule; correct order of evaluation (not outside interval 120 to 195); 12.5 OE with no sight of premature approximation |
| Alternative: \(AC^2 = (26\sin\theta)^2 + (31.5-26\cos\theta)^2 = 10^2 + 7.5^2\), \(AC=12.5\) | (M1)(m1)(A1) |
# Question 2:
## Part (a)
$\text{Area} = \frac{1}{2}(26)(31.5)\sin\theta = \frac{1}{2}(26)(31.5)\times\frac{5}{13} = 157.5 \text{ cm}^2$ | M1, A1 | $\frac{1}{2}(26)(31.5)\sin(\theta)$ stated or used; OE e.g. $\frac{315}{2}$; Condone AWRT 157.50; NMS: 157.5 or AWRT 157.50 scores B2
## Part (b)
$\cos\theta = \frac{12}{13}$ | B1 | $\frac{12}{13}$ OE exact fraction
## Part (c)
$AC^2 = 31.5^2 + 26^2 - 2\times31.5\times26\times\cos(\theta)$
$= 992.25 + 676 - 1512 = 1668.25 - 1512 = 156.25$
$AC = \sqrt{156.25} = 12.5\text{ cm}$ | M1, m1, A1 | RHS of cosine rule; correct order of evaluation (not outside interval 120 to 195); 12.5 OE with no sight of premature approximation
Alternative: $AC^2 = (26\sin\theta)^2 + (31.5-26\cos\theta)^2 = 10^2 + 7.5^2$, $AC=12.5$ | (M1)(m1)(A1) |
---2 The triangle $A B C$, shown in the diagram, is such that $A B = 26 \mathrm {~cm}$ and $B C = 31.5 \mathrm {~cm}$.
The acute angle $A B C$ is $\theta$, where $\sin \theta = \frac { 5 } { 13 }$.
\begin{enumerate}[label=(\alph*)]
\item Calculate the area of triangle $A B C$.
\item Find the exact value of $\cos \theta$.
\item Calculate the length of $A C$.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2012 Q2 [6]}}