Triangle with Ratio of Sides Equal to Ratio of Angles In an equilateral triangle, the side lengths are in ratio 1:1:1, as are the angle measures.
Are there also non-equilateral triangles in which the ratio of the side lengths is the same as the ratio of the angle measures? If so, what is an example of such a triangle?
 A: The equilateral triangle is the only such case.
Let $A$ and $B$ be two angles, and $a$ and $b$ be the respective opposite sides. Then 
$$
\frac{\sin A}{\sin B} = \frac a b
$$
so if your ratios match, then
$$
\frac{\sin A}{\sin B} = \frac a b
$$
$$
\frac{\sin A}{A} = \frac{\sin B}{B} 
$$
So to have a triangle where $A \neq B$ you require two distinct positive points, both  on the curve $f(x) = \frac{\sin x}x$ , having the same value of $f(x)$.  For this to happen one of the points must be greater than $\pi$ and that cannot be an angle in a triangle.
So no two angles of such a triangle can be unequal, thus it must be equilateral.
A: Using the Law of Sines, the ratio of sides to the ratio of angles, becomes the ratio of sines of angles to the ratio of angles. This means that the sinc of the angles must be equal. Since the Sinc Function is one-to-one on $[0,\pi]$, the angles must be equal.
Therefore, the only triangle to have this property is the equilateral triangle.

That is, we want
$$
\frac a\alpha=\frac b\beta=\frac c\gamma\tag{1}
$$
The Law of Sines says
$$
\frac{\sin(\alpha)}a=\frac{\sin(\beta)}b=\frac{\sin(\gamma)}c\tag{2}
$$
Multiplying $(1)$ and $(2)$, we get
$$
\frac{\sin(\alpha)}\alpha=\frac{\sin(\beta)}\beta=\frac{\sin(\gamma)}\gamma\tag{3}
$$
Since the sinc function is one-to-one on $[0,\pi]$, $(3)$ implies that
$$
\alpha=\beta=\gamma\tag{4}
$$
