What if you were given a 30-60-90 right triangle and the length of one of its side? How could you figure out the lengths of its other sides? After completing this Concept, you'll be able to use the 30-60-90 Theorem to solve problems like this one.

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CK-12 Foundation: Chapter8306090RightTrianglesA

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James Sousa: Trigonometric Function Values of Special Angles

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James Sousa: Solving Special Right Triangles

### Guidance

One of the special right triangles is called a 30-60-90 triangle, after its three angles. 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*} and the hypotenuse is \begin{align*}2x\end{align*}. Use the Pythagorean Theorem to find the longer leg. What is it? How is this similar to your answer in #4?

\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*}, the longest leg is always \begin{align*}x \sqrt{3}\end{align*}, and the hypotenuse is always \begin{align*}2x\end{align*}. If you ever forget this corollary, then you can still use the Pythagorean Theorem.

#### Example A

Find the length of the missing sides.

We are given the shortest leg. If \begin{align*}x = 5\end{align*}, then the longer leg, \begin{align*}b = 5 \sqrt{3}\end{align*}, and the hypotenuse, \begin{align*}c = 2(5) = 10\end{align*}.

#### Example B

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

#### Example C

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.

Watch this video for help with the Examples above.

CK-12 Foundation: Chapter8306090RightTrianglesB

### Guided Practice

1. Find the length of the missing sides.

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

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*}.

**Answers:**

1. 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*}.

2. 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*}

3. 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*}

### Explore More

- 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)?

### Answers for Explore More Problems

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