(b) ADB and BCD are right-angled triangles. BC = CD. AD = \(10\sqrt{6}\) mm. Angle BAD = 30°.
\(\tan 30^\circ = \dfrac{1}{\sqrt{3}}\)
Not to scale
Work out the length of BC. [6]
Mark scheme (a)
Answer
Marks
Part marks and guidance
\(\frac{1}{\sqrt{2}}\) or \(\frac{\sqrt{2}}{2}\) final answer
1
Mark scheme (b)
Answer
Marks
Part marks and guidance
10 nfww
6
B3 for BD = \(10\sqrt{2}\) oe or M2 for \(10\sqrt{6} \times \tan 30\) oe or M1 for \(\frac{BD}{10\sqrt{6}} = \tan 30\) oe AND M2 for BC = \(\sqrt{\frac{\textit{their}\ BD^2}{2}}\) oe or their BD × their sin 45 oe or M1 for BC2 + CD2 = (their BD)2 or for \(\frac{BC}{\textit{their}\ BD} = \textit{their}\ \sin 45\)
Allow use of other variables for BC and CD (possibly different)
