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$\text{What is the ratio of legs in a right triangle with angles of 15, 75, and 90?}$ 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.

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    So now you have $1:(\sqrt 3+2):\sqrt{1^2+(\sqrt 3+2)^2}$ – 2017-01-03

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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.)

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    And I might add: That is a very well-reasoned picture, and good thinking to come up with that answer without any reference to trigonometry. – 2017-01-03
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    Yes. On the other hand, the picture is a nice and easy way to explain *why* $\tan 15^\circ$ equals $2-\sqrt3$ exactly. – 2017-01-03
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    @Alex Thanks a lot for this and I liked how you actually proved that the tan 15 $= 2-\sqrt{3}$. :) – 2017-01-05
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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).

$${a\over b}=\tan(15^\circ)\quad\text{or}\quad{b\over a}=\tan(75^\circ)$$