Ratio of Legs in 15, 75, 90 triangles $\text{What is the ratio of legs in a right triangle with angles of 15, 75, and 90?}$ I know the ratio of legs in a $30, 60, 90$ triangle, which is the lengths $1$, $\sqrt{3}$, and $2$ respectively. This is what I have got so far:
Using the 30-60-90 Ratio
How would I be able to take this a step further and be able to find the answer? Thanks in advance.
 A: The ratio of legs is
$$
r = \tan 15^\circ.
$$
(This is quite easily derived from the definition of the $\tan$ function.)
You can also represent the ratio using radicals:
$$
r = 2 - \sqrt{3} \approx 0.267949
$$
If we do not want to use $\tan$ at all, then we obtain the same answer just reasoning from your picture:
$$
r = {1\over2+\sqrt{3}}= 2 - \sqrt{3}.
$$
(In this ratio, the numerator $1$ is the vertical leg in your picture; and the denominator $2+\sqrt{3}$ is the horizontal leg.)
A: Let $a$, $b$, $c$ represent the line segments the make up a right triangle, and let $A$, $B$, $C$ represent the angles opposite those line segments. Let $C=90^\circ$
The legs ($a$ and $b$) are given by the following equations.
$$a=c\sin{A}$$
$$b=c\cos{A}$$
We want to find the ratio $a\over b$.
$${a\over b}={c\sin{A}\over c\cos{A}}={\sin{A}\over \cos{A}}=\tan(A)$$
Now substitute the value of $A$ in your example. You can either use $15^\circ$ or $75^\circ$ (using one instead of the other will give the inverse ratio).
$${a\over b}=\tan(15^\circ)\quad\text{or}\quad{b\over a}=\tan(75^\circ)$$
A: See the link here, it provides an explanation of the ratios and has other resources on triangle ratios:
https://robertlovespi.net/2013/10/18/the-15-75-90-triangle/
Brief overview:
15-75-90 triangle is based on the regular dodecagon, using this diagram:
Using the Pythagoran Theorem and other methods, we get the short leg:long leg:hypotenuse ratio in a 15-75-90 triangle as $2-\sqrt{3}:1:2\sqrt{2 - \sqrt{3}}.$ The nested radical ultimately simplifies to $\sqrt{6} - \sqrt{2},$ so the simplified ratio is $2 - \sqrt{3}:1:\sqrt{6} - \sqrt{2}.$
