| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Point on side of triangle |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard applications of the cosine rule, triangle area formula, and sine rule. Part (i) is direct cosine rule application, part (ii) uses the standard area formula (1/2)ab sin C, and part (iii) requires sine rule in triangle ABD. All are routine techniques with no problem-solving insight required, making it slightly 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 | Guidance |
|---|---|---|
| (i) \(BC = 9^2 + 17^2 - 2 \times 9 \times 17 \times \cos 40°\) with \(BC = 11.6\) cm | M1 | Attempt use of correct cosine rule. Must be correct formula seen or implied, but allow a slip when evaluating eg omission of 2, or incorrect use of an additional big bracket. Allow M1 even if subsequently evaluated in radian mode (23.96). Allow M1 if expression is not square rooted, as long as it is clear that correct formula was used ie either \(BC^2 = ...\). or even just \(a^2 = ...\) if the power disappears from \(BC\). |
| Obtain 11.6, or better | A1 | 2 |
| (ii) \(\text{area} = \frac{1}{2} \times 9 \times 17 \times \sin 40° = 49.2 \text{ cm}^2\) | M1 | Attempt triangle area using \((\frac{1}{2})ab\sin C\), or equiv. Condone omission of \(\frac{1}{2}\) from this formula, but no other errors allowed. If using right-angled triangle, must use \(\frac{1}{2}bh\) with reasonable attempt at perpendicular sides. Allow M1 if subsequently evaluated in radian mode (57.00). If using 40°, must be using sides of 9 and 17, not 11.6 from (i). If using another angle, can still get M1 as long as sides used are consistent with this angle. |
| Obtain 49.2, or better | A1 | 2 |
| (iii) \(\frac{BD}{9} = \frac{\sin 40}{\sin 63}\) with \(BD = 6.49\) cm | M1 | Attempt use of correct sine rule, or equiv, to find length \(BD\). No further rearrangement required. Could have both fractions the other way up. Must be angles of 40° and 63° if finding \(BD\) directly. Must be attempting \(BD\), so using 77° to find \(AD\) is M0 unless attempt is then made to find \(BD\) by any valid method. Placing \(D\) on \(BC\) is M0. |
| Obtain correct unsimplified expression involving \(BD\) as the only unknown | A1 | Can still get A1 even if evaluated in radians (40.07). If using a multi-step method (eg use 77° to find \(AD\) and then use cosine rule to find \(BD\)) then this A mark is only given when a correct (unsimplified) expression involving \(BD\) as the only unknown is obtained. |
| Obtain 6.49, or better | A1 | 3 |
**(i)** $BC = 9^2 + 17^2 - 2 \times 9 \times 17 \times \cos 40°$ with $BC = 11.6$ cm | M1 | Attempt use of correct cosine rule. Must be correct formula seen or implied, but allow a slip when evaluating eg omission of 2, or incorrect use of an additional big bracket. Allow M1 even if subsequently evaluated in radian mode (23.96). Allow M1 if expression is not square rooted, as long as it is clear that correct formula was used ie either $BC^2 = ...$. or even just $a^2 = ...$ if the power disappears from $BC$.
**Obtain 11.6, or better** | A1 | 2 | Actual answer is 11.644329..., so allow more accurate answer as long as it rounds to 11.64
**(ii)** $\text{area} = \frac{1}{2} \times 9 \times 17 \times \sin 40° = 49.2 \text{ cm}^2$ | M1 | Attempt triangle area using $(\frac{1}{2})ab\sin C$, or equiv. Condone omission of $\frac{1}{2}$ from this formula, but no other errors allowed. If using right-angled triangle, must use $\frac{1}{2}bh$ with reasonable attempt at perpendicular sides. Allow M1 if subsequently evaluated in radian mode (57.00). If using 40°, must be using sides of 9 and 17, not 11.6 from (i). If using another angle, can still get M1 as long as sides used are consistent with this angle.
**Obtain 49.2, or better** | A1 | 2 | Actual answer is 49.17325..., so allow more accurate answer as long as it rounds to 49.17. Must come from correct working only.
**(iii)** $\frac{BD}{9} = \frac{\sin 40}{\sin 63}$ with $BD = 6.49$ cm | M1 | Attempt use of correct sine rule, or equiv, to find length $BD$. No further rearrangement required. Could have both fractions the other way up. Must be angles of 40° and 63° if finding $BD$ directly. Must be attempting $BD$, so using 77° to find $AD$ is M0 unless attempt is then made to find $BD$ by any valid method. Placing $D$ on $BC$ is M0.
**Obtain correct unsimplified expression involving $BD$ as the only unknown** | A1 | Can still get A1 even if evaluated in radians (40.07). If using a multi-step method (eg use 77° to find $AD$ and then use cosine rule to find $BD$) then this A mark is only given when a correct (unsimplified) expression involving $BD$ as the only unknown is obtained.
**Obtain 6.49, or better** | A1 | 3 | Actual answer is 6.492756..., so allow more accurate answer as long as it rounds to 6.493. Must come from correct working only not eg sin 117.
---1
The diagram shows triangle $A B C$, with $A B = 9 \mathrm {~cm} , A C = 17 \mathrm {~cm}$ and angle $B A C = 40 ^ { \circ }$.\\
(i) Find the length of $B C$.\\
(ii) Find the area of triangle $A B C$.\\
(iii) $D$ is the point on $A C$ such that angle $B D A = 63 ^ { \circ }$. Find the length of $B D$.
\hfill \mbox{\textit{OCR C2 2011 Q1 [7]}}