June 2017 Paper 1 Q14
14 In the diagram, \(BCD\) is a straight line.
\(AD = 2\sqrt{3}\) cm
Not drawn accurately
Work out the exact length of \(CD\).
Give your answer in the form \(\quad a + b\sqrt{3} \quad\) where \(a\) and \(b\) are integers. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(AB = \sqrt{3}\) | B1 | |
| Any one of these responses ... \(\dfrac{BD}{2\sqrt{3}} = \cos 30^\circ \qquad \dfrac{BD}{2\sqrt{3}} = \sin 60^\circ\) \(\dfrac{\sqrt{3}}{BD} = \tan 30^\circ \qquad \dfrac{BD}{\sqrt{3}} = \tan 60^\circ\) \(BD^2 + (\sqrt{3})^2 = (2\sqrt{3})^2\) oe | M1 | ... or these ... \(\dfrac{BD}{2\sqrt{3}} = \dfrac{\sqrt{3}}{2} \qquad \dfrac{BD}{\sqrt{3}} = \sqrt{3} \qquad \dfrac{\sqrt{3}}{BD} = \dfrac{1}{\sqrt{3}}\) \(\dfrac{BD}{\sin 60^\circ} = \dfrac{\sqrt{3}}{\sin 30^\circ} \qquad \dfrac{BD}{\sqrt{3}/2} = \dfrac{\sqrt{3}}{1/2}\) |
| \(BD = 3\) | A1 | |
| \(CD = 3 - \sqrt{3}\) | A1 | oe |
Additional guidance
SC1 for a final answer of \(\dfrac{2\sqrt{3}\sin 15^\circ}{\sin 135^\circ}\), possibly with \(\dfrac{1}{\sqrt{2}}\) or \(\dfrac{\sqrt{2}}{2}\) for \(\sin 135^\circ\)