| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Perpendicular from vertex |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard applications of the cosine rule, area formula (½ab sin C), and perpendicular height calculation. Part (a) is routine cosine rule application with a 'show that' requiring minimal work. Parts (b)(i) and (b)(ii) are direct formula applications with no problem-solving insight needed. Slightly easier than average due to the guided structure and standard techniques. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos\theta\) | M1 | Correct use of cosine rule |
| \(100 = 25 + 64 - 80\cos\theta\) | A1 | Correct substitution |
| \(\cos\theta = \frac{-11}{80}\) | A1 | \(\theta = 97.9°\) (shown) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Area \(= \frac{1}{2}(5)(8)\sin(97.9°)\) | M1 | Correct formula with their \(\theta\) |
| \(= 19.8 \text{ cm}^2\) | A1 | 3 significant figures |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{Area} = \frac{1}{2} \times BC \times AD\) | M1 | Using area with base BC |
| \(19.8... = \frac{1}{2} \times 10 \times AD\) | M1 | Correct equation |
| \(AD = 3.96 \text{ cm}\) | A1 | 3 significant figures |
# Question 3:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos\theta$ | M1 | Correct use of cosine rule |
| $100 = 25 + 64 - 80\cos\theta$ | A1 | Correct substitution |
| $\cos\theta = \frac{-11}{80}$ | A1 | $\theta = 97.9°$ (shown) |
## Part (b)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Area $= \frac{1}{2}(5)(8)\sin(97.9°)$ | M1 | Correct formula with their $\theta$ |
| $= 19.8 \text{ cm}^2$ | A1 | 3 significant figures |
## Part (b)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{Area} = \frac{1}{2} \times BC \times AD$ | M1 | Using area with base BC |
| $19.8... = \frac{1}{2} \times 10 \times AD$ | M1 | Correct equation |
| $AD = 3.96 \text{ cm}$ | A1 | 3 significant figures |
---3 The triangle $A B C$, shown in the diagram, is such that $A B = 5 \mathrm {~cm} , A C = 8 \mathrm {~cm}$, $B C = 10 \mathrm {~cm}$ and angle $B A C = \theta$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\theta = 97.9 ^ { \circ }$, correct to the nearest $0.1 ^ { \circ }$.
\item \begin{enumerate}[label=(\roman*)]
\item Calculate the area of triangle $A B C$, giving your answer, in $\mathrm { cm } ^ { 2 }$, to three significant figures.
\item The line through $A$, perpendicular to $B C$, meets $B C$ at the point $D$. Calculate the length of $A D$, giving your answer, in cm , to three significant figures.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2011 Q3 [8]}}