#### 30-60-90 Tirnalges

One of the special right triangles is called a 30-60-90 triangle, after its three angles.

*Watch the first half of this video.*

To construct a 30-60-90 triangle, start with an equilateral triangle.

#### Investigation: Properties of a 30-60-90 Triangle

Tools Needed: Pencil, paper, ruler, compass

1. Construct an equilateral triangle with 2 in sides.

2. Draw or construct the altitude from the top vertex to the base for two congruent triangles.

3. Find the measure of the two angles at the top vertex and the length of the shorter leg.

*The top angles are each* \begin{align*}30^\circ\end{align*}*and the shorter leg is 1 in because the altitude of an equilateral triangle is also the angle and perpendicular bisector.*

4. Find the length of the longer leg, using the Pythagorean Theorem. Simplify the radical.

5. Now, let’s say the shorter leg is length \begin{align*}x\end{align*}

\begin{align*}x^2 + b^2 & = (2x)^2\\
x^2 + b^2 & = 4x^2\\
b^2 & = 3x^2\\
b & = x \sqrt{3}\end{align*}

**30-60-90 Corollary:** If a triangle is a 30-60-90 triangle, then its sides are in the extended ratio \begin{align*}x : x \sqrt{3} : 2x\end{align*}

Step 5 in the above investigation proves the 30-60-90 Corollary. The shortest leg is always \begin{align*}x\end{align*}

#### Finding the Length of Missing Sides

Find the length of the missing sides.

We are given the shortest leg. If \begin{align*}x = 5\end{align*}

#### Solving for Unknown Values

Find the value of \begin{align*}x\end{align*}

We are given the longer leg.

\begin{align*}x \sqrt{3} &= 12\\
x &= \frac{12}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} \\ x&= \frac{12 \sqrt{3}}{3} \\x&= 4 \sqrt{3}\end{align*}

Then, the hypotenuse would be \begin{align*}y = 2 \left ( 4 \sqrt{3} \right ) = 8 \sqrt{3}\end{align*}.

*Watch the first half of this video*.

#### Finding Unknown Lengths

Find the measure of \begin{align*}x\end{align*}.

Think of this trapezoid as a rectangle, between a 45-45-90 triangle and a 30-60-90 triangle.

From this picture, \begin{align*}x = a + b + c\end{align*}. First, find \begin{align*}a\end{align*}, which is a leg of an isosceles right triangle.

\begin{align*}a = \frac{24}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{24 \sqrt{2}}{2} = 12 \sqrt{2}\end{align*}

\begin{align*}a = d\end{align*}, so we can use this to find \begin{align*}c\end{align*}, which is the shorter leg of a 30-60-90 triangle.

\begin{align*}c = \frac{12 \sqrt{2}}{ \sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{12 \sqrt{6}}{3} = 4 \sqrt{6}\end{align*}

\begin{align*}b = 20\end{align*}, so \begin{align*}x = 12 \sqrt{2} + 20 + 4 \sqrt{6}\end{align*}. Nothing simplifies, so this is how we leave our answer.

### Examples

#### Example 1

Find the length of the missing sides.

We are given the hypotenuse. \begin{align*}2x = 20\end{align*}, so the shorter leg, \begin{align*}f = 10\end{align*}, and the longer leg, \begin{align*}g = 10 \sqrt{3}\end{align*}.

#### Example 2

Find the value of \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.

We are given the hypotenuse.

\begin{align*}2x &= 15 \sqrt{6}\\ x &= \frac{15 \sqrt{6}}{2}\\ \end{align*}

The, the longer leg would be \begin{align*}y = \left ( \frac{15 \sqrt{6}}{2} \right ) \cdot \sqrt{3} = \frac{15 \sqrt{18}}{2} = \frac{45 \sqrt{2}}{2}\end{align*}

#### Example 3

\begin{align*}x\end{align*} is the hypotenuse of a 30-60-90 triangle and \begin{align*}y\end{align*} is the longer leg of the same triangle. The shorter leg has a length of \begin{align*}6\end{align*}.

We are given the shorter leg.

\begin{align*}& x=2(6)\\ & x = 12\\ & \text{The longer leg is}\\ & y= 6 \cdot \sqrt{3} = 6 \sqrt{3}\end{align*}

### Review

- In a 30-60-90 triangle, if the shorter leg is 5, then the longer leg is __________ and the hypotenuse is ___________.
- In a 30-60-90 triangle, if the shorter leg is \begin{align*}x\end{align*}, then the longer leg is __________ and the hypotenuse is ___________.
- A rectangle has sides of length 6 and \begin{align*}6 \sqrt{3}\end{align*}. What is the length of the diagonal?
- Two (opposite) sides of a rectangle are 10 and the diagonal is 20. What is the length of the other two sides?

For questions 5-12, find the lengths of the missing sides. Simplify all radicals.

- What is the height of an equilateral triangle with sides of length 3 in?
- What is the area of an equilateral triangle with sides of length 5 ft?
- A regular hexagon has sides of length 3 in. What is the area of the hexagon? (
*Hint: the hexagon is made up a 6 equilateral triangles.*) - The area of an equilateral triangle is \begin{align*}36 \sqrt{3}\end{align*}. What is the length of a side?
- If a road has a grade of \begin{align*}30^\circ\end{align*}, this means that its angle of elevation is \begin{align*}30^\circ\end{align*}. If you travel 1.5 miles on this road, how much elevation have you gained in feet (5280 ft = 1 mile)?

### Review (Answers)

To view the Review answers, open this PDF file and look for section 8.5.