| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Sequential triangle calculations (basic) |
| Difficulty | Easy -1.2 This is a straightforward three-part question testing basic sine/cosine rule applications with simple numbers (3, 4, 30°). Part (a) uses the standard area formula, (b) applies cosine rule directly, and (c) uses sine rule - all routine procedures with no problem-solving required, making it easier than average. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C) |
| Answer | Marks |
|---|---|
| \(\frac{1}{2} \times 3 \times 4 \times \sin 30\) | M1 |
| \(= 3\) | A1 |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(AC^2 = 3^2 + 4^2 - 2\times3\times4\times\cos 30°\) \((= 4.22)\) | M1 | or \(AC = \sqrt{3^2 + 4^2 - 2\times3\times4\times\cos 30°}\) Correct expression \(AC^2\) or \(AC\) |
| \(AC = 2.05\) (3 sf) | A1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{\sin C}{3} = \frac{\sin 30}{\text{their } 2.05}\) oe eg \(\frac{\sin C}{3} = \frac{\sin 30}{\sqrt{4.22}}\) | M1 | Correct sin rule using their (b) |
| \(ACB = \sin^{-1}\!\left(\frac{3\sin 30}{\text{their } 2.05}\right)\) \((= \sin^{-1}0.73\ldots)\) | M1 | Attempt inverse sine of \(\frac{3\sin 30}{\text{their(b)}}\). May be implied by answer |
| \(3^2 = \text{'2.05'}^2 + 4^2 - 2\times\text{'2.05'}\times4\times\cos ACB\) | M1 | Correct cos rule using their (b) |
| \(ACB = \cos^{-1}\frac{\text{'2.05'}^2+4^2-3^2}{2\times\text{'2.05'}\times4}\) | M1 | Attempt inverse cos of \(\frac{\text{'2.05'}^2+4^2-3^2}{2\times\text{'2.05'}\times4}\). May be implied by answer |
| \(ACB = 46.9°\) or \(47.0°\) or \(47.1°\) (3 sf) | A1ft | Allow 47°. Condone premature rounding. FT their (b) |
| [3] |
# Question 1:
## Part (a):
| $\frac{1}{2} \times 3 \times 4 \times \sin 30$ | M1 | |
|---|---|---|
| $= 3$ | A1 | |
| | [2] | |
## Part (b):
| $AC^2 = 3^2 + 4^2 - 2\times3\times4\times\cos 30°$ $(= 4.22)$ | M1 | or $AC = \sqrt{3^2 + 4^2 - 2\times3\times4\times\cos 30°}$ Correct expression $AC^2$ or $AC$ |
|---|---|---|
| $AC = 2.05$ (3 sf) | A1 | |
| | [2] | |
## Part (c):
| $\frac{\sin C}{3} = \frac{\sin 30}{\text{their } 2.05}$ oe eg $\frac{\sin C}{3} = \frac{\sin 30}{\sqrt{4.22}}$ | M1 | Correct sin rule using their (b) |
|---|---|---|
| $ACB = \sin^{-1}\!\left(\frac{3\sin 30}{\text{their } 2.05}\right)$ $(= \sin^{-1}0.73\ldots)$ | M1 | Attempt inverse sine of $\frac{3\sin 30}{\text{their(b)}}$. May be implied by answer |
| $3^2 = \text{'2.05'}^2 + 4^2 - 2\times\text{'2.05'}\times4\times\cos ACB$ | M1 | Correct cos rule using their (b) |
| $ACB = \cos^{-1}\frac{\text{'2.05'}^2+4^2-3^2}{2\times\text{'2.05'}\times4}$ | M1 | Attempt inverse cos of $\frac{\text{'2.05'}^2+4^2-3^2}{2\times\text{'2.05'}\times4}$. May be implied by answer |
| $ACB = 46.9°$ or $47.0°$ or $47.1°$ (3 sf) | A1ft | Allow 47°. Condone premature rounding. FT their (b) |
| | [3] | |
---1 In the triangle $A B C , A B = 3 , B C = 4$ and angle $A B C = 30 ^ { \circ }$. Find the following.
\begin{enumerate}[label=(\alph*)]
\item The area of the triangle.
\item The length $A C$.
\item The angle $A C B$.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q1 [7]}}