2.12: Exact tan 15° and Pythagoras

0606/21/O/N/20 — Question 6 · 5 marks

Do not use a calculator in this question. All lengths are in centimetres.

2.12 diagram
In the diagram, AC=3−1AC = \sqrt{3} - 1, AB=3+1AB = \sqrt{3} + 1, ∠ABC=15°\angle ABC = 15^{\degree} and ∠CAB=90°\angle CAB = 90^{\degree}.
(a)
Show that tan⁡15°=2−3\tan 15^{\degree} = 2 - \sqrt{3}.
[3]
(b)
Find the exact length of BCBC.
[2]