One angle in a triangle is twice the other, find the relationship among the sides

Question

$\angle A$ is twice as $\angle B$. I need to find formula that describes relationship among $a,b,c$. It should be a function of only $a,b,c$.

I tried using cosine law and sine law but no result.

Answer 1

$$\frac{a}{b}=\frac{\sin A }{\sin B}={2\cos B } \tag1$$

which is independent $c $

Apply Cosine Rule and make $c$ to be $B$ dependent:

$$ c^2= a^2+b^2- 2 a b \cos (\pi- 3B)=a^2+b^2+ 2 a b \cos (3B)\tag2$$

Now you know $$ \cos 3B = 4 \cos^3 B -3 \cos B \tag3$$ so eliminate the trig. functions to take it to the next.

EDIT1:

Simplify

$$ \frac{c^2-a^2-b^2}{a}= \frac{a^3}{b^2} -3a \tag4$$

resulting in two simple relations:

$$ a^2-b^2 - bc = 0 ,\quad a^2-b^2+ bc=0 \tag5 $$

This gives

$$ \pm c = \frac{a^2}{b}-b \tag6 $$

A check with right triangle $ (A,B,C)=( 60^{\circ} ,30^{\circ} ,90 ^{\circ}) $ decides that the second relation is to be discarded. $ [2= (\sqrt3)^2-1 ]$

$$ c = \frac{a^2}{b}-b \tag7 $$

which is the required relation. The relation is also checked with geometric construction with angles $ (A,B,C)=( 40^{\circ} ,80^{\circ} ,60 ^{\circ}). $

Answer 2

$$\cos\alpha=\cos2\beta$$ or $$\sin\alpha=\sin2\beta,$$ which gives $$\frac{b^2+c^2-a^2}{2bc}=2\left(\frac{a^2+c^2-b^2}{2ac}\right)^2-1$$ or $$\frac{a}{b}=\frac{\sin\alpha}{\sin\beta}=\frac{\sin2\beta}{\sin\beta}=2\cos\beta=\frac{a^2+c^2-b^2}{ac}.$$