June 2018 Paper 1 Q20
20 In triangle \(ABC\),
\(AB = 6\sqrt{2}\) cm, angle \(ABC = 45^\circ\) and angle \(ACB = 60^\circ\)
Work out the value of \(x\).
Give your answer in the form \(a\sqrt{b}\), where \(a\) and \(b\) are integers.
You must show your working. [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(\sin 45 = \dfrac{1}{\sqrt{2}}\) or \(\dfrac{\sqrt{2}}{2}\) or \(\sin 60 = \dfrac{\sqrt{3}}{2}\) | B1 | look for these in their working … one of them correct will gain the mark at any stage of their working |
| \(\dfrac{x}{\sin 45} = \dfrac{6\sqrt{2}}{\sin 60}\) | M1 | oe |
| \((x =)\ \dfrac{6\sqrt{2} \times \frac{1}{\sqrt{2}}}{\frac{\sqrt{3}}{2}}\) or \(\dfrac{6\sqrt{2} \times \frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{2}}\) | M1dep | oe a correct expression or equation coming from both surds correct eg \(\ \dfrac{\sqrt{3}}{2}x = 6\sqrt{2} \times \dfrac{1}{\sqrt{2}}\) or \(\dfrac{6}{x} = \dfrac{\sqrt{3}}{2}\) |
| \((x =)\ \dfrac{12}{\sqrt{3}}\) or \(\sqrt{3}x = 12\) | A1 | |
| \((x =)\ 4\sqrt{3}\) | A1 | |
| Alternative method 2 | ||
| \(\sin 45 = \dfrac{1}{\sqrt{2}}\) or \(\dfrac{\sqrt{2}}{2}\) or \(\sin 60 = \dfrac{\sqrt{3}}{2}\) | B1 | look for these in their working … one of them correct will gain the mark at any stage of their working |
| \(\dfrac{AD}{6\sqrt{2}} = \sin 45\) or \((AD =)\ 6\sqrt{2} \times \sin 45\) or \((AD =)\ 6\) | M1 | where D is the foot of the perpendicular from A to BC |
| \((x =)\ \dfrac{6\sqrt{2} \times \frac{1}{\sqrt{2}}}{\frac{\sqrt{3}}{2}}\) or \(\dfrac{6\sqrt{2} \times \frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{2}}\) | M1dep | oe a correct expression or equation coming from both surds correct eg \(\ \dfrac{\sqrt{3}}{2}x = 6\sqrt{2} \times \dfrac{1}{\sqrt{2}}\) or \(\dfrac{6}{x} = \dfrac{\sqrt{3}}{2}\) |
| \((x =)\ \dfrac{12}{\sqrt{3}}\) or \(\sqrt{3}x = 12\) | A1 | |
| \((x =)\ 4\sqrt{3}\) | A1 | |
Additional guidance
| If one surd value is incorrect they can only score a maximum of 2 marks |