Answer: \(\tan15^\circ=2-\sqrt3\); \(BC=2\sqrt2\).

(a) In the right-angled triangle,

\(\tan15^\circ=\frac{AC}{AB}.\)

Using the given lengths,

\(\tan15^\circ=\frac{\sqrt3-1}{\sqrt3+1}.\)

Rationalise the denominator:

\(\frac{\sqrt3-1}{\sqrt3+1}\times\frac{\sqrt3-1}{\sqrt3-1} =\frac{(\sqrt3-1)^2}{3-1}.\)

Now

\((\sqrt3-1)^2=3-2\sqrt3+1=4-2\sqrt3.\)

Therefore

\(\tan15^\circ=\frac{4-2\sqrt3}{2}=2-\sqrt3.\)

(b) By Pythagoras' theorem,

\(BC^2=(\sqrt3-1)^2+(\sqrt3+1)^2.\)

Now

\((\sqrt3-1)^2=4-2\sqrt3\)

and

\((\sqrt3+1)^2=4+2\sqrt3.\)

So

\(BC^2=8.\)

Hence

\(BC=\sqrt8=2\sqrt2.\)