Higher June 2022 Paper 5 Q15
15 ABC is an equilateral triangle of side length 2 cm.
M is the midpoint of AC.
Not to scale
Using this diagram, show that \(\tan 30^\circ = \dfrac{1}{\sqrt{3}}\). [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
Correctly shows \(\tan 30^\circ = \frac{1}{\sqrt{3}}\) with supporting working | 4 | Using triangle MBC (mark similarly for use of triangle MBA): B1 for MBC = 30° or MC = 1 [cm] soi by use in Pythagoras M2 for [BM =] \(\sqrt{2^2 - 1^{[2]}}\) or \(\sqrt{4 - 1}\) [\(= \sqrt{3}\)] or for [BM2 =] \(2^2 - 1^{[2]}\) oe and \(\sqrt{3}\) or M1 for \(\text{BM}^2 + 1^{[2]} = 2^2\) oe OR B1 for MCB = 60° or MC = 1 [cm] soi by use in cos rule M2 for \(\sqrt{2^2 + 1^{[2]} - 2 \times 2 \times 1 \times \cos 60}\) and \(\cos 60 = \frac{1}{2}\) soi [\(= \sqrt{3}\)] or M1 for \(2^2 + 1^{[2]} - 2 \times 2 \times 1 \times \cos 60\) OR B1 for MCB = 60° soi by use in sine ratio or sine rule M2 for [\(x\) = ] 2sin 60 oe and \(\sin 60 = \frac{\sqrt{3}}{2}\) soi [\(= \sqrt{3}\)] or M1 for \(\frac{\sin 60}{x} = \frac{\sin 90}{2}\) oe AND A1dep for MBC = 30° stated/on diagram and \(\tan 30^\circ = \frac{1}{\sqrt{3}}\) and with no errors seen leading to the answer | If 0 or 1 scored, SC2 for \(\dfrac{\sin 30}{\cos 30} = \dfrac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\) and \(\tan 30 = \frac{1}{\sqrt{3}}\) May be on diagram For M2 must show the subtraction, do not allow \(\text{BM}^2 + 1^{[2]} = 2^2\) and then \(\sqrt{3}\), this gets M1 only M2 accept [\(x\) = ] \(\dfrac{2\sin 60}{\sin 90}\) and \(\sin 60 = \frac{\sqrt{3}}{2}\) Dep on B1M2 A0 for \(\tan = \frac{1}{\sqrt{3}}\) (without the 30) |