| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sine and Cosine Rules |
| Type | Quadrilateral with diagonal |
| Difficulty | Standard +0.3 This is a standard multi-part sine/cosine rule question requiring systematic application of learned techniques: (i) cosine rule in triangle ABD, (ii) cosine rule in triangle BCD to find an angle, (iii) area calculation using two triangles. While it has multiple steps (6-7 marks typical), each step follows a predictable pattern with no novel insight required, making it slightly easier than the average A-level question. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05c Area of triangle: using 1/2 ab sin(C) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(BD^2 = 6^2 + 9^2 - (2 \times 6 \times 9 \times \cos 60)\) | M1 | |
| \(BD^2 = 36 + 81 - 54 = 63\) | ||
| \(BD = \sqrt{63} = \sqrt{9 \times 7} = 3\sqrt{7}\) cm | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \((3\sqrt{7})^2 = 3^2 + 8^2 - (2 \times 3 \times 8 \times \cos C)\) | M1 | |
| \(\cos C = \frac{9 + 64 - 63}{48} = \frac{5}{24}\) | M1 | |
| \(\angle BCD = 78.0°\) (1dp) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(= (\frac{1}{2} \times 6 \times 9 \times \sin 60) + (\frac{1}{2} \times 3 \times 8 \times \sin 77.975)\) | M2 | |
| \(= 35.1\) cm² (3sf) | A1 | (9) |
# Question 7:
## Part (i):
| Answer/Working | Marks | Notes |
|---|---|---|
| $BD^2 = 6^2 + 9^2 - (2 \times 6 \times 9 \times \cos 60)$ | M1 | |
| $BD^2 = 36 + 81 - 54 = 63$ | | |
| $BD = \sqrt{63} = \sqrt{9 \times 7} = 3\sqrt{7}$ cm | M1 A1 | |
## Part (ii):
| Answer/Working | Marks | Notes |
|---|---|---|
| $(3\sqrt{7})^2 = 3^2 + 8^2 - (2 \times 3 \times 8 \times \cos C)$ | M1 | |
| $\cos C = \frac{9 + 64 - 63}{48} = \frac{5}{24}$ | M1 | |
| $\angle BCD = 78.0°$ (1dp) | A1 | |
## Part (iii):
| Answer/Working | Marks | Notes |
|---|---|---|
| $= (\frac{1}{2} \times 6 \times 9 \times \sin 60) + (\frac{1}{2} \times 3 \times 8 \times \sin 77.975)$ | M2 | |
| $= 35.1$ cm² (3sf) | A1 | **(9)** |
---7.
The diagram shows the quadrilateral $A B C D$ in which $A B = 6 \mathrm {~cm} , B C = 3 \mathrm {~cm}$, $C D = 8 \mathrm {~cm} , A D = 9 \mathrm {~cm}$ and $\angle B A D = 60 ^ { \circ }$.\\
(i) Show that $B D = 3 \sqrt { 7 } \mathrm {~cm}$.\\
(ii) Find the size of $\angle B C D$ in degrees to 1 decimal place.\\
(iii) Find the area of quadrilateral $A B C D$.\\
\hfill \mbox{\textit{OCR C2 Q7 [9]}}