What are the angles of an isosceles triangle with a specific ratio?

[anemone]

Let $ABC$ be an isosceles triangle such that $AB=AC$. Find the angles of $\triangle ABC$ if $\dfrac{AB}{BC}=1+2\cos\dfrac{2\pi}{7}$.

 

[Opalg]

Spoiler

Write $\alpha$ for the two equal angles in the isosceles triangle, so that the angle at the apex is $\pi - 2\alpha$. By the sine rule, $$\frac{AB}{AC} = \frac{\sin\alpha}{\sin(\pi - 2\alpha)} = \frac{\sin\alpha}{\sin( 2\alpha)} = \frac{\sin\alpha}{2\sin\alpha\cos\alpha} = \frac1{2\cos\alpha}.$$ Now for a bit of trigonometry: $$\begin{aligned} \sin(3\theta) = \sin(2\theta+\theta) &= \sin(2\theta)\cos\theta + \cos(2\theta)\sin\theta \\ &= 2\sin\theta\cos^2\theta + \cos(2\theta)\sin\theta = \sin\theta(2\cos^2\theta + \cos(2\theta)) = \sin\theta(1 + 2\cos(2\theta)) \end{aligned}$$ (because $2\cos^2\theta = \cos(2\theta) + 1$). Therefore $1+ 2\cos(2\theta) = \dfrac{\sin(3\theta)}{\sin\theta}.$ In particular, with $\theta = \frac\pi7$, $$1 + 2\cos\tfrac{2\pi}7 = \frac{\sin\frac{3\pi}7}{\sin\frac{\pi}7} = \frac{\sin\frac{3\pi}7}{\sin\frac{6\pi}7} = \frac{\sin\frac{3\pi}7}{2\sin\frac{3\pi}7\cos\frac{3\pi}7} = \frac1{2\cos\frac{3\pi}7}.$$ It follows that if \(\displaystyle \frac{AB}{AC} = 1 + 2\cos\tfrac{2\pi}7\) then \(\displaystyle \frac1{2\cos\alpha} = \frac1{2\cos\frac{3\pi}7}\), so that $\alpha = \frac{3\pi}7$. Thus the angles of the triangle are $\frac{3\pi}7$, $\frac{3\pi}7$ and $\frac\pi7$.

 

[anemone]

Bravo, Opalg!(Cool)