October 2021 Paper 3 Q11
11 In this question you must show detailed reasoning.
The diagram shows triangle ABC, with BC = 8 cm and angle BAC = 45°. The point D on AC is such that DC = 5 cm and BD = 7 cm.
Determine the exact length of AB. [5]
| Scheme | Marks | AO |
|---|---|---|
| DR In triangle BDC, \(\cos D = \dfrac{7^2 + 5^2 - 8^2}{2 \times 7 \times 5}\) | M1 | 3.1a |
| \(\cos D = \frac{1}{7}\) | A1 | 1.1 |
| \(\sin D = \sqrt{1 - \frac{1}{49}}\) \(\sin D = \dfrac{\sqrt{48}}{7}\) | M1 | 1.1 |
| \(\dfrac{\mathrm{AB}}{\sin D} = \dfrac{7}{\sin 45^\circ} \Rightarrow \dfrac{7\mathrm{AB}}{\sqrt{48}} = \dfrac{7 \times 2}{\sqrt{2}}\) | M1 | 3.1a |
| \(\mathrm{AB} = 4\sqrt{6}\) [cm] oe | A1 | 2.2a |
| [5] |
Notes
DR: This question included the instruction: In this question you must show detailed reasoning.
M1: Use of cosine rule in triangle BDC (for any angle)
Or \(\cos C = \dfrac{5^2 + 8^2 - 7^2}{2 \times 5 \times 8}\)
A1: Or \(\cos C = \frac{1}{2}\) Or \(C = 60^\circ\)
M1: Approximate values must not be seen to earn M1 i.e. must be exact
Or \(\sin C = \dfrac{\sqrt{3}}{2}\)
M1: Use of sin rule in triangle ABD
Exact values must be seen to earn M1
Or use of sin rule in triangle ABC \(\dfrac{2\mathrm{AB}}{\sqrt{3}} = \dfrac{8 \times 2}{\sqrt{2}}\)
A1: Must be exact answer