# What Are 30°–60°–90° Triangles?

When you have a triangle with the angles $30$°, $60$°, and $90$°, the length of the shorter leg is always half the length of the hypotenuse—the longest side, longer than the two legs. You also know that the longer leg is always equal to $\sqrt{3}$ multiplied by the length of the shorter leg.

Rule

In a “$30$°-$60$°-$90$°” triangle, the sides have the following relationships:

where $\sqrt{3}\approx 1.73$.

Example 1

A $\text{}30\text{}\text{°}$-$\text{}60\text{}\text{°}$-$\text{}90\text{}\text{°}$ triangle has a hypotenuse of length $\text{}12\text{}\phantom{\rule{0.17em}{0ex}}\text{cm}$. Find the length of the legs.

Since this is a “$30$°-$60$°-$90$°” triangle, you already know that the shorter leg measures half the length of the hypotenuse, or $k=\frac{12\phantom{\rule{0.17em}{0ex}}\text{cm}}{2}=6\phantom{\rule{0.17em}{0ex}}\text{cm}$. You can now find the lengtt of the longer leg either by using the Pythagorean theorem, or by taking $\sqrt{3}k$. Here I’ll show you both of them.

First, the Pythagorean theorem: $\begin{array}{llll}\hfill {k}^{2}+{6}^{2}& =1{2}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {k}^{2}& =144-36\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \sqrt{{k}^{2}}& =\sqrt{108}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill k& \approx 10.39\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

With $\sqrt{3}k$:

The longer leg is about $10.39$ cm long.

Example 2

You have a $\text{}30\text{}\text{°}$-$\text{}60\text{}\text{°}$-$\text{}90\text{}\text{°}$ triangle where the longer leg measures $\text{}3\text{}\phantom{\rule{0.17em}{0ex}}\text{cm}$. Find the length of the two other sides.

Since this is a “$30$°-$60$°-$90$°” triangle, you know that the shorter leg’s length is half of the hypotenuse $h$. Therefore you can make the following equation using the Pythagorean theorem: $\begin{array}{llll}\hfill {3}^{2}+\phantom{\rule{-0.17em}{0ex}}{\left(\frac{1}{2}h\right)}^{2}& ={h}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 9& ={h}^{2}-\frac{1}{4}{h}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 9& =\frac{3}{4}\cdot {h}^{2}\phantom{\rule{2em}{0ex}}|\cdot \frac{4}{3}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \frac{4}{3}\cdot 9& ={h}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \sqrt{12}& =\sqrt{{h}^{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 3.5& \approx h\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The hypotenuse measures about $3.5$ cm. The shorter leg is half of the hypotenuse, so it is

 $k\approx \frac{3.5\phantom{\rule{0.17em}{0ex}}\text{cm}}{2}\approx 1.75\phantom{\rule{0.17em}{0ex}}\text{cm}$

Example 3

You have an equilateral triangle where all sides are $\text{}6\text{}\phantom{\rule{0.17em}{0ex}}\text{cm}$. What is the height of the triangle?

Because the angles in an equilateral triangle are all $60$°, you can find the height of the triangle by using the formula for a “$30$°-$60$°-$90$°” triangle. The height becomes one of the legs in a “$30$°-$60$°-$90$°” triangle:

As you can see, the hypotenuse is $6$ cm, and the shorter leg of the new triangle is exactly half of the hypotenuse (and half of the length of the side of the original triangle), which is $3$ cm.

You then calculate: $\begin{array}{llll}\hfill {h}^{2}+{3}^{2}& ={6}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {h}^{2}& =36-9\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \sqrt{{h}^{2}}& =\sqrt{27}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill h& \approx 5.2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The height $h$ is therefore about $5.2$ cm.