# How do you find the remaining side of a 30^circ-60^circ-90^circ triangle if the side opposite 60^circ is 4?

Aug 6, 2017

$\frac{4 \sqrt{3}}{3}$ and $\frac{8 \sqrt{3}}{3}$

#### Explanation:

A ${30}^{\circ} - {60}^{\circ} - {90}^{\circ}$ triangle is a type of special right triangle because the side lengths always occur in the ratio $x - x \sqrt{3} - 2 x$.

The ${30}^{\circ}$ angle is opposite to the side length $x$, the ${60}^{\circ}$ angle is opposite to $x \sqrt{3}$, and the ${90}^{\circ}$ angle is opposite to $2 x$, as shown below.

If the side opposite to ${60}^{\circ}$ is $4$, then $x \sqrt{3} = 4$. To find the remaining sides, first solve for $x$:

$x \sqrt{3} = 4$

$x = \frac{4}{\sqrt{3}}$ $\to$ divide both sides by $\sqrt{3}$

$x = \frac{4}{\sqrt{3}} \cdot \textcolor{red}{\frac{\sqrt{3}}{\sqrt{3}}}$ $\to$ rationalize the denominator by multiplying by $\frac{\sqrt{3}}{\sqrt{3}}$, which is equal to $1$

$x = \frac{4 \sqrt{3}}{3}$ $\to$ multiply

Lastly, to find the side opposite to the ${90}^{\circ}$ angle, find $2 x$:

$2 x = 2 \cdot \frac{4 \sqrt{3}}{3} = \frac{8 \sqrt{3}}{3}$

So, the sides are $\frac{4 \sqrt{3}}{3} , 4 ,$ and $\frac{8 \sqrt{3}}{3}$.