| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | January |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Sequential triangle calculations (basic) |
| Difficulty | Moderate -0.8 This is a straightforward two-part application of the sine rule requiring no problem-solving insight. Part (i) uses sine rule to find an angle given two sides and an angle, part (ii) then finds the remaining side. Both are direct, single-step calculations that are easier than the typical multi-technique A-level question. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{\sin A}{10} = \frac{\sin 63}{14}\) | M1 | Attempt use of correct sine rule; must be correct sine rule, either way up. Need to rearrange at least as far as \(\sin A = ...\), using a valid method. Allow M1 even if subsequently evaluated in radians (0.120) |
| \(A = 39.5°\) | A1 | Obtain \(39.5°\), or better. Actual answer is 39.52636581... so allow more accurate answer as long as it rounds to 39.53 |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(c^2 = 10^2 + 14^2 - 2 \times 10 \times 14 \times \cos 77.5°\) | M1 | Attempt use of correct cosine rule, or equiv, inc attempt at \(77.5°\). Angle used must be \(77.5°\) or must come from a clear attempt at \(180-(63+\text{their } A)\). NB Using \(102.5°\) in sine rule will give 15.3, but this is M0. Must be correct formula seen or implied, but allow slip when evaluating e.g. omission of 2, incorrect extra 'big bracket'. Allow M1 if expression is not square rooted, as long as LHS was intended to be correct ie \(c^2 = ...\) or \(AB^2 = ...\). Allow M1 even if subsequently evaluated in rad mode. Allow any equiv method, including sine rule (as far as \(\sin C = ...\)) or right-angled triangle trig (must be full and valid method) |
| \(c = 15.3\) | A1 | Obtain 15.3, or better. Allow more accurate answer as long as it rounds to 15.34 |
| [2] |
# Question 1:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{\sin A}{10} = \frac{\sin 63}{14}$ | M1 | Attempt use of correct sine rule; must be correct sine rule, either way up. Need to rearrange at least as far as $\sin A = ...$, using a valid method. Allow M1 even if subsequently evaluated in radians (0.120) |
| $A = 39.5°$ | A1 | Obtain $39.5°$, or better. Actual answer is 39.52636581... so allow more accurate answer as long as it rounds to 39.53 |
| | [2] | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $c^2 = 10^2 + 14^2 - 2 \times 10 \times 14 \times \cos 77.5°$ | M1 | Attempt use of correct cosine rule, or equiv, inc attempt at $77.5°$. Angle used must be $77.5°$ or must come from a clear attempt at $180-(63+\text{their } A)$. **NB** Using $102.5°$ in sine rule will give 15.3, but this is M0. Must be correct formula seen or implied, but allow slip when evaluating e.g. omission of 2, incorrect extra 'big bracket'. Allow M1 if expression is not square rooted, as long as LHS was intended to be correct ie $c^2 = ...$ or $AB^2 = ...$. Allow M1 even if subsequently evaluated in rad mode. Allow any equiv method, including sine rule (as far as $\sin C = ...$) or right-angled triangle trig (must be full and valid method) |
| $c = 15.3$ | A1 | Obtain 15.3, or better. Allow more accurate answer as long as it rounds to 15.34 |
| | [2] | |
---1
The diagram shows triangle $A B C$, with $A C = 14 \mathrm {~cm} , B C = 10 \mathrm {~cm}$ and angle $A B C = 63 ^ { \circ }$.\\
(i) Find angle $C A B$.\\
(ii) Find the length of $A B$.
\hfill \mbox{\textit{OCR C2 2013 Q1 [4]}}