How were the values of this trigonmetric ratio determined? I'm reading a book that is pretty spartan about definitions. How did the book come up with the length of the sides of this triangle?

I understand the trig ratios once we have the lengths... but how were the lengths of $\sqrt{3}$, 1, and 2  determined? I think the book is assuming that the radius is 2 and that the terminal angle (unsure if this is the right word... but the angle created by the terminal side and the x-axis) is $\frac{\pi}{3}$. But how did we get the other two sides that are not the assumed radius?
 A: It's a geometric property. Suppose you have a $\triangle ABC$ with right angle and $C$ and let $\angle ABC=\frac{\pi}{6}$. Then $AB=2\cdot AC$
To see why this is true, let $M$ be the midpoint of AB. Then as $AM=BM=CM$ and $\angle BAC=\frac{\pi}{3}$ $\triangle AMC$ is equilateral, hence $AC=AM=\frac{AB}{2}$. You can then determine $BC$ by using the Pythagorean Theorem.
Hence, for such triangles, specifying one of the sides' length let's you determine the length of the other sides.
A: A bit of a story.
In 8th grade (mid-1980s for me), my math teacher drilled us on five Pythagorean triplets:


*

*$1, 1, \sqrt{2}$

*$1, \sqrt{3}, 2$

*$3,4,5$

*$5,12,13$

*$8,15,17$


We learned to find the missing side(s) by pattern-matching for these triangles. (We had a dozen triangles, and we were given four minutes to finish the quiz. There wasn't time to calculate using $a^2 + b^2 = c^2$.)
The motivation was that the state math exams used these triplets all the time on the questions, and so becoming familiar with these triplets made those questions a piece of cake.
So, when I see $1, \sqrt{3}, 2$ or even $1/2, \sqrt{3}/2, 1$, I just know, cold, that it's a $30-60-90$ triangle, and everything else just falls into place. Perhaps it's that familiarity that the textbook writer is counting on. Or, that it's cleaner not to have fractions for the sides.
Anyway, it's a very common example for trigonometry. If you're looking for a casting of the triangle to the unit circle, then you can divide each side by $2$, so the hypotenuse is $1$.
A: One of the angles of a right triangle is $60^\circ.$ If you believe that the sum of the three angles is $180^\circ,$ then the third one is $30^\circ.$
So take a mirror image of that triangle, with the vertical side as the mirror, and put those two triangles together, and you have a triangle in which all three angles are $60^\circ.$ So all three sides have the same length, which is $2.$ One of those three sides consists of the shortest side of the triangle depicted and the shortest side of its mirror image. Therefore the length of that shortest side is $1.$
Now use the Pythagorean theorem to find the length of the other leg of the right triangle: $$\sqrt{2^2 - 1^2} = \sqrt 3 \, .$$
How you know that the sum of the angles of every triangle is $180^\circ$ is similarly simple, and you might want to look at that too.
