| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2009 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Bearings and navigation |
| Difficulty | Standard +0.3 This is a straightforward application of the sine rule in part (i), followed by routine use of the cosine rule in part (ii), and finding perpendicular distance in part (iii). All steps are standard C2 techniques with no novel 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 |
|---|---|---|
| \(\frac{TA}{\sin 107} = \frac{50}{\sin 3}\) | M1 | Attempt use of correct sine rule to find \(TA\), or equiv |
| \(TA = 914\) m | A1 2 | Obtain 914, or better |
| Answer | Marks | Guidance |
|---|---|---|
| \(TC = \sqrt{914^2 + 150^2 - 2 \times 914 \times 150 \times \cos 70}\) | M1, A1\(\sqrt{}\) | Attempt use of correct cosine rule; correct unsimplified expression following (i) |
| \(= 874\) m | A1 3 | Obtain 874, or better |
| Answer | Marks | Guidance |
|---|---|---|
| dist from \(A = 914 \times \cos 70 = 313\) m; beyond \(C\), hence 874 m is shortest dist | M1, A1 2 | Attempt to locate point of closest approach; convincing argument point is beyond \(C\), or obtain 859, or better |
| OR perp dist \(= 914 \times \sin 70 = 859\) m | SR B1 for 874 stated with no method shown |
# Question 5:
## Part (i):
| $\frac{TA}{\sin 107} = \frac{50}{\sin 3}$ | M1 | Attempt use of correct sine rule to find $TA$, or equiv |
|---|---|---|
| $TA = 914$ m | A1 **2** | Obtain 914, or better |
## Part (ii):
| $TC = \sqrt{914^2 + 150^2 - 2 \times 914 \times 150 \times \cos 70}$ | M1, A1$\sqrt{}$ | Attempt use of correct cosine rule; correct unsimplified expression following (i) |
|---|---|---|
| $= 874$ m | A1 **3** | Obtain 874, or better |
## Part (iii):
| dist from $A = 914 \times \cos 70 = 313$ m; beyond $C$, hence 874 m is shortest dist | M1, A1 **2** | Attempt to locate point of closest approach; convincing argument point is beyond $C$, or obtain 859, or better |
|---|---|---|
| **OR** perp dist $= 914 \times \sin 70 = 859$ m | | **SR** B1 for 874 stated with no method shown |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{bbee5a50-4a32-4171-8713-8eb38914a511-3_623_355_1123_897}
Some walkers see a tower, $T$, in the distance and want to know how far away it is. They take a bearing from a point $A$ and then walk for 50 m in a straight line before taking another bearing from a point $B$. They find that angle $T A B$ is $70 ^ { \circ }$ and angle $T B A$ is $107 ^ { \circ }$ (see diagram).\\
(i) Find the distance of the tower from $A$.\\
(ii) They continue walking in the same direction for another 100 m to a point $C$, so that $A C$ is 150 m . What is the distance of the tower from $C$ ?\\
(iii) Find the shortest distance of the walkers from the tower as they walk from $A$ to $C$.
\hfill \mbox{\textit{OCR C2 2009 Q5 [7]}}