Triangle Swivel

Here's a bit of a hacky solution. It's not complete because it assumes that the general form provided is correct. However, I think it's kind of nice.

When $\theta = \frac{\pi}{6}$ , the area of the small triangle is zero:

$\implies cos(\frac{\pi}{6} + \frac{\pi}{\beta}) = 0 \implies \beta = 3$

When $\theta = 0$ , the area of the small triangle is the same as the area of the big triangle:

$\implies \alpha \, cos^2(\frac{\pi}{3}) = 1 \\ \alpha \frac{1}{4} = 1 \implies \alpha = 4$

Knowing that the angles of an equilateral triangle are each 60°, we can find the other angles easily: Now we can use the sin rule to get the side lengths x and y in terms of r and theta: $\frac{a}{sinA}=\frac{b}{sinB}$ $\begin{aligned}\frac{a}{\sin A}&=\frac{b}{\sin B}&\frac{a}{\sin A}&=\frac{b}{\sin B}\\ \frac{x}{\sin(60^{\circ}-\theta)}&=\frac{r}{\sin(120^{\circ})}&\frac{y}{\sin(\theta)}&=\frac{r}{\sin(120^{\circ})}\\ x&=\frac{2r\sin(60^{\circ}-\theta)}{\sqrt3}&y&=\frac{2r\sin(\theta)}{\sqrt3}\\ x&=\frac{2r(\sin60^{\circ}\cos\theta-\cos60^{\circ}\sin\theta)}{\sqrt3}\\ x&=\frac{r(\sqrt3\cos\theta-\sin\theta)}{\sqrt3}\end{aligned}$ The side length of the small triangle is: $\begin{aligned}x-y&=\frac{r}{\sqrt3}(\sqrt3\cos\theta-3\sin\theta)\\ &=r(\cos\theta-\sqrt3\sin\theta)\end{aligned}$ Now is a good time to use auxiliary angle method to reduce it to: $x-y=2r\cos\left(\theta+\frac{\pi}{3}\right)$ And finally, we use the formula for the area of a triangle: $Area\,of\,a\,triangle=\frac{1}{2}ab\sin C$ $\begin{aligned}Area\,of\,small\,triangle&=\frac{1}{2}\left(2r\cos\left(\theta+\frac{\pi}{3}\right)\right)^2\sin60^{\circ}&Area\,of\,big\,triangle&=\frac{1}{2}r^2\sin60^{\circ}\\ &=\sqrt3r^2\cos^2\left(\theta+\frac{\pi}{3}\right)&&=\frac{\sqrt3}{4}r^2\end{aligned}$ $\begin{aligned}\frac{Area\,of\,small\,triangle}{Area\,of\,big\,triangle}&=\frac{\sqrt3r^2\cos^2\left(\theta+\frac{\pi}{3}\right)}{\frac{\sqrt3}{4}r^2}\\ &=4\cos^2\left(\theta+\frac{\pi}{3}\right)\end{aligned}$ So $\alpha=4$ and $\beta=3$ , so the answer is 7.