Does congruent triangles apply to this question?

Question

Two identical rods $BA$ and $CA$ are hinged at $A$. When $BC = 8\ \textrm{cm}$, $\angle BAC = 30^\circ$ and when $BC = 4\ \textrm{cm}$, $\angle BAC = \alpha$. Show that $$\cos\alpha = \frac{6+\sqrt 3}{8}$$

I drew two diagrams and tried finding the length of $AC$ (which is the same as $AB$). Can I use that to solve the triangle with $BC=4$ using cosine rule?

Answer 1

Let $AB=AC=l$. The cosine rule in case of 30 degrees gives $$ l^2+l^2-2l^2\cos 30^\circ=8^2. $$ Since $\cos 30^\circ=\sqrt3/2$, this is $$ l^2=\frac{8^2}{2-\sqrt3}. $$ Now the case of $BC=4$ gives $$ \cos\alpha=\frac{2l^2-4^2}{2l^2}=1-\frac8{l^2}=\dots $$ as desired.

Answer 2

$\cos$ rule will be quite lengthy. Use basic trigonometry instead. Let the length of rod be $AB=AC=l$. Now, $$\sin 15°=\frac{4}{l}\tag{1}$$ and $$\sin\frac{\alpha}{2}=\frac{2}{l}\tag{2}$$ I think you can solve now.