The perimeter of an isosceles triangle $\triangle ABC$ 
An isosceles triangle $\triangle ABC$ is given with $\angle ACB=30^\circ$ and leg $BC=16$ $cm$. Find the perimeter of $\triangle ABC$.

We have two cases, right? When 1) $AC=BC=16$ and 2) $AB=BC=16$.
For the first case: let $CH$ be the altitude through $C$. Since the triangle is isosceles, $CH$ is also the angle bisector and $\measuredangle ACH=\measuredangle BCH=15^\circ$. How to approach the problem further? I have not studied trigonometry.

The first case.
Let $BK$ be an altitude of $\Delta ABC$.
Thus, $$BK=8,$$
$$CK=\sqrt{BC^2-BK^2}=\sqrt{16^2-8^2}=8\sqrt3$$ and
$$AB=\sqrt{AK^2+BK^2}=\sqrt{(16-8\sqrt3)^2+8^2}=$$
$$=\sqrt{8^2(2-\sqrt3)^2+8^2}=8\sqrt{(2-\sqrt3)^2+1}=16\sqrt{2-\sqrt3},$$
which gives the answer: $32+16\sqrt{2-\sqrt3}.$
We can solve the second problem by the similar way.
