30-60-90直角三角形：边之比

A special type of right triangle is the $30$-$60$-$90$ right triangle. As you can tell from the name of the triangle, this right triangle has interior angles of $30^{\circ}$, $60^{\circ}$, and $90^{\circ}$.

Unlike the 45-45-90 right triangle, the side ratios of the triangle are not as straightforward. We need to prove the ratio of the length of the shortest leg to the length of the hypotenuse before we are able to determine all the side ratios.

Let us create a diagram in which we reflect the $30$-$60$-$90$ triangle along the longer leg. Since the two $90^{\circ}$ angles fit together on the same line, we have created a larger triangle - that resembles an equilateral triangle!

We can easily prove that it is an equilateral triangle by noting our larger triangle's angles. One of the angles of our original $30$-$60$-$90$ right triangle is $60^{\circ}$, so there are two $60^{\circ}$ angles. The third angle is also $60^{\circ}$ because it is the sum of two $30^{\circ}$ angles.

Suppose the length of the shortest side of one $30$-$60$-$90$ triangle is $1$ unit. Therefore, the length of a side of the equilateral triangle would be $2$ units!

This means that the ratio of the lengths of the shortest side to the hypotenuse of any $30$-$60$-$90$ right triangle is $1:2$.

If the length of the shortest leg is $a$ units and the hypotenuse is $c$ units, we can use the Pythagorean Theorem to derive the length of the longer leg, denoted as $b$: $$\begin{align} c^2 &= a^2 + b^2 \\ b^2 &= c^2 - a^2 \\ & = (2a)^2 - a^2 \\ & = 4a^2 - a^2 \\ & = 3a^2 \\ \therefore b & = \sqrt{3}a\ \text{units} \end{align}$$

Therefore,

If a triangle is a $30$-$60$-$90$ right triangle, the ratio of the sides (short leg:long leg:hypotenuse) is $1:\sqrt{3}:2$.

(Video) $30$-$60$-$90$ Triangles

In the video below, Richard Rusczyk and, uh, Harvey show you how to find the legs of a $30$-$60$-$90$ triangle. Spoiler alert: you can treat $30$-$60$-$90$ like two halves of an equilateral triangle!

If you say that the shortest leg of a $30$-$60$-$90$ triangle has length $x$, the rest of the legs will always follow the ratio:

$$\text{short leg : long leg : hypotenuse}\\ x \,\,\,\,\, : \,\,\,\,\, x\sqrt{3} \,\,\,\,\, : \,\,\,\,\, 2x$$

If you want to see Richard work through figuring out the ratio specifically, just start the video at timestamp 5:10! After proving the ratio, Richard also solves a problem by just using what he knows about 30-60-90 triangles (even though the problem is about a rectangle).

First off: draw a picture!

Next, let's see what we can figure out from the information in the problem. We know that triangle $ADC$ has a $30°$ and a $90°$ angle, so it must be a $30$-$60$-$90$ triangle. And the same goes for triangle $ABC$! Let's update our picture to reflect that:

Now that we know we're working with two $30$-$60$-$90$ triangles, we can use the ratios to figure out more information about our picture!

Since the long leg of a $30$-$60$-$90$ triangle $\sqrt{3}$ times the length of the shortest leg, we can figure out that line $AC$ has a length of $x\sqrt{3}$:

Now we know the length of the shortest leg of triangle $ABC$! And because we know the ratio, that means that we know the length of the hypotenuse--the length of line $AB$.

$$2 \times x\sqrt{3} = 2x\sqrt{3}$$

And that's our answer!