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# 30-60-90 Right Triangles

## Hypotenuse equals twice the smallest leg, while the larger leg is sqrt(3) times the smallest.

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30-60-90 Triangles
As Lindsay was walking near the Civic Center she noticed a shadow on the ground. She looked at the shadow and then cast her eyes upward to see the top of a flagpole. “That flagpole is quite tall but I bet I had to look up a longer distance than the pole is tall,” Lindsay said to herself. “I would like to figure out the length of my line of sight from the ground to the top of the flagpole,” she added.

How can Lindsay use the diagram below to figure out the length of her line of sight?

In this concept, you will learn about \begin{align*}30^\circ-60^\circ-90^\circ\end{align*} right triangles.

### Equilateral Triangle

An equilateral triangle is one that has three sides of equal length and three angles of equal measure. The angles of the triangle each measure 60°. When an altitude is drawn in an equilateral triangle, it bisects the angle from which it is drawn and is a perpendicular bisector of the base. A perpendicular bisector of the base passes through the midpoint of the base and is perpendicular to that side. The first diagram shows an equilateral triangle and the second one shows an equilateral triangle and its altitude.

In the second triangle the line \begin{align*}\overline{CM}\end{align*} is the altitude of the equilateral triangle. The measure of \begin{align*}\angle C\end{align*} has been divided into two angles each measuring 30°. The altitude meets the base of the triangle \begin{align*}\overline{AB}\end{align*} at right angles and divides the base into \begin{align*}\overline{AM}\end{align*} and \begin{align*}\overline{MB}\end{align*} which are equal in length. If \begin{align*}\overline{AM}+\overline{MB}=\overline{AB}\end{align*} then the length of \begin{align*}\overline{AB}=x+x=2x\end{align*}. If \begin{align*}\overline{AB}=\overline{AC}=\overline{BC}\end{align*} then \begin{align*}\overline{AC}=2x\end{align*} and \begin{align*}\overline{BC}=2x\end{align*}.

The altitude of the equilateral triangle has divided it into two equal right triangles. \begin{align*}\Delta BCM\end{align*} is a \begin{align*}30^\circ-60^\circ-90^\circ\end{align*} triangle and it will be used to show the relationship between the legs and the hypotenuse of this special right triangle.

Let’s begin by using the Pythagorean Theorem to find the measure of the altitude \begin{align*}\overline{CM}\end{align*}.

First, write down the values of \begin{align*}(a,b,c)\end{align*} for the Pythagorean Theorem.

\begin{align*} \begin{array}{rcl} \overline{CM} &=& a \ = \ ? \\ \overline{MB} &=& b \ = \ x \\ \overline{BC} &=& c \ = \ 2x \end{array}\end{align*}

Next, write the Pythagorean Theorem and substitute in the values for \begin{align*}(a,b,c)\end{align*}.

\begin{align*}\begin{array}{rcl} c^2 &=& a^2+b^2 \\ (2x)^2 &=& a^2+(x)^2 \end{array}\end{align*}

Next, do the indicated squaring on both sides of the equation.

\begin{align*}\begin{array}{rcl} (2x)^2 &=& a^2+(x)^2 \\ (2x \cdot 2x) &=& a^2+(x \cdot x) \\ 4x^2 &=& a^2+x^2 \end{array}\end{align*}

Next, isolate the variable by subtracting \begin{align*}x^2\end{align*} from both sides of the equation and simplify.

\begin{align*} \begin{array}{rcl} 4x^2 &=& a^2+x^2 \\ 4x^2 - x^2 &=& a^2+x^2-x^2 \\ 3x^2 &=& a^2 \end{array}\end{align*}

Then, solve for the variable ‘\begin{align*}a\end{align*}’ by taking the square root of both sides of the equation.

\begin{align*}\begin{array}{rcl} 3x^2 &=& a^2 \\ \sqrt{3} \cdot \sqrt{x^2} &=& \sqrt{a^2} \\ \sqrt{3} \cdot x &=& a \\ x \sqrt{3} &=& a \end{array}\end{align*}

The answer is \begin{align*}x \sqrt{3}\end{align*}.

If \begin{align*}b=x\end{align*} then the answer can be expressed as \begin{align*}b \sqrt{3}\end{align*}.

The relationship between the sides of the \begin{align*}30^\circ-60^\circ-90^\circ\end{align*} triangle is:

• The length of the hypotenuse is twice the length of the shorter leg.
• The length of the shorter leg is one-half the length of the hypotenuse.
• The length of the longer leg is \begin{align*}\sqrt{3}\end{align*} times the length of the shorter leg.

### Examples

#### Example 1

Earlier, you were given a problem about Lindsay and the flagpole shadow. She wants to know the length of her line of sight to the top of the flagpole.

Lindsay will have to use the relationships between the sides of a special triangle.

First, she should write down what she knows.

The length of the flagpole is the longer leg of the \begin{align*}30^\circ-60^\circ-90^\circ\end{align*} triangle and is \begin{align*}7 \sqrt{3} \ \text{feet}\end{align*} in length.

The length of the shadow is the shorter leg and its length is unknown.

The length of the hypotenuse is the length of Lindsay’s line of sight and she wants to figure out this length.

Next, identify the variables.

Let ‘\begin{align*}b\end{align*}’ represent the length of the longer leg, ‘\begin{align*}a\end{align*}’ represent the length of the shorter leg and ‘\begin{align*}c\end{align*}’ represent her line of sight.

Next, write down an equation showing the relationship between the two legs of the triangle.

\begin{align*}b=\sqrt{3} \cdot a\end{align*}

Next, fill any known values into the equation.

\begin{align*}\begin{array}{rcl} b &=& \sqrt{3} \cdot a \\ 7 \sqrt{3} &=& a \sqrt{3} \end{array}\end{align*}

Then, divide both sides of the equation by \begin{align*}\sqrt{3}\end{align*} to solve for the variable ‘\begin{align*}a\end{align*}’.

\begin{align*}\begin{array}{rcl} 7 \sqrt{3} &=& a \sqrt{3} \\ \frac{7 \cancel{ \sqrt{3}}}{ \cancel{ \sqrt{3}}} &=& \frac{a \cancel{ \sqrt{3}}}{\cancel{ \sqrt{3}}}\\ 7 &=& a \end{array}\end{align*}

The length of the shorter leg is 7 feet.

Next, write an equation showing the relationship between the hypotenuse and the shorter leg.

\begin{align*}c=2a\end{align*}

Next, substitute the length of the shorter leg into the equation.

\begin{align*}\begin{array}{rcl} c &=& 2a \\ c &=& 2(7) \end{array}\end{align*}

Then, perform the multiplication on the right side of the equation.

\begin{align*}\begin{array}{rcl} c &=& 2(7) \\ c &=& 14 \end{array}\end{align*}

The length of Lindsay’s line of sight is 14 feet.

#### Example 2

For the following \begin{align*}30^\circ-60^\circ-90^\circ\end{align*} triangle with a hypotenuse of 16 inches, find the length of the longer leg to the nearest hundredth.

First, write down what you know from the diagram.

The hypotenuse is \begin{align*}\overline{BC}=16 \ \text{inches}\end{align*}

The shorter leg is \begin{align*} \overline{AC}\end{align*}.

The longer leg is \begin{align*}\overline{AB}\end{align*}.

Next, write an equation that shows the relationship between the shorter leg and the hypotenuse.

\begin{align*}\overline{AC}= \frac{1}{2} (\overline{BC})\end{align*}

Then, substitute the value for \begin{align*}\overline{BC}\end{align*} into the equation and simplify.

\begin{align*}\begin{array}{rcl} \overline{AC} &=& \frac{1}{2} (\overline{BC}) \\ \overline{AC} &=& \frac{1}{2} (16) \\ \overline{AC} &=& 8 \end{array}\end{align*}

The length of the shorter leg is 8 inches.

Now determine the length of the longer leg \begin{align*}\overline{AB}\end{align*}.

First, write an equation that shows the relationship between the two legs of the triangle.

\begin{align*}\overline{AB} = \overline{AC} \cdot \sqrt{3}\end{align*}

Next, substitute the value of \begin{align*}\overline{AC}\end{align*} into the equation and simplify.

\begin{align*}\begin{array}{rcl} \overline{AB} &=& \overline{AC} \cdot \sqrt{3} \\ \overline{AB} &=& (8) \cdot \sqrt{3} \\ \overline{AB} &=& 8 \sqrt{3} \end{array}\end{align*}

Then, use the calculator to express the length of the longer leg to the nearest hundredth.

\begin{align*}\begin{array}{rcl} \overline{AB} &=& 8 \sqrt{3} \\ \overline{AB} &=& 8(1.73) \\ \overline{AB} &\approx& 13.8 \end{array}\end{align*}

The length of the longer leg is approximately 13.8 inches.

#### Example 3

For the following \begin{align*}30^\circ-60^\circ-90^\circ\end{align*} triangle with a longer leg of 6 inches, find the length of the shorter leg. Express the length as a radical.

First, write down what you know from the diagram.

The longer leg is \begin{align*}\overline{CD}=6 \ \text{inches}\end{align*}

The shorter leg is \begin{align*}\overline{EC}\end{align*}.

The hypotenuse is \begin{align*}\overline{ED}\end{align*}.

Next, write an equation that shows the relationship between the longer leg and the shorter leg.

\begin{align*}\overline{CD}= \sqrt{3}(\overline{EC})\end{align*}

Next, substitute the value for \begin{align*}\overline{CD}\end{align*} into the equation and simplify.

\begin{align*} \begin{array}{rcl} \overline{CD} &=& \sqrt{3}(\overline{EC}) \\ 6 &=& \sqrt{3}(\overline{EC}) \end{array}\end{align*}

Next, divide both sides of the equation by \begin{align*}\sqrt{3}\end{align*} to solve \begin{align*}\overline{EC}\end{align*}

\begin{align*}\begin{array}{rcl} 6 &=& \sqrt{3}(EC) \\ \frac{6}{\sqrt{3}} &=& \frac{ \cancel {\sqrt{ 3}} (EC)}{ \cancel {\sqrt{ 3}}} \\ \frac{6}{\sqrt{3}} &=& \overline{EC} \end{array}\end{align*}

Next, multiply the left side of the equation by \begin{align*}\frac{\sqrt{3}}{\sqrt{3}}\end{align*} to clear the radical from the denominator.

\begin{align*}\begin{array}{rcl} \frac{6}{ \sqrt{3}} &=& \overline{EC} \\ \frac{6}{ \sqrt{3}} \cdot \left( \frac{ \sqrt{3}}{ \sqrt{3}} \right ) &=& \overline{EC} \\ \frac{6 \sqrt{3}}{ \sqrt{9}}&=& \overline{EC} \end{array}\end{align*}
Next, simplify the left side of the equation.
\begin{align*} \begin{array}{rcl} \frac{6 \sqrt{3}}{ \sqrt{9}}&=& \overline{EC} \\ \frac{6 \sqrt{3}}{ 3}&=& \overline{EC} \\ \frac{\overset{2}{\cancel{6} \sqrt{3}}}{\cancel{3}} &=& \overline{EC} \\ 2 \sqrt{3}&=& \overline{EC} \end{array} \end{align*}

The answer is \begin{align*}2 \sqrt{3}\end{align*}.

The length of the shorter leg is \begin{align*}2 \sqrt{3}\end{align*} inches.

#### Example 4

For the following \begin{align*}30^\circ-60^\circ-90^\circ\end{align*} triangle with a shorter leg of 4 feet, find the length of the hypotenuse to the nearest tenth.

First, write down what you know from the diagram.

The hypotenuse is \begin{align*}\overline{AB} \ = \ ? \ \text{feet}\end{align*}

The shorter leg is \begin{align*}\overline{AC} \ = \ 4 \ \text{feet}\end{align*}

The longer leg is \begin{align*}\overline{BC}\end{align*}.

Next, write an equation that shows the relationship between the shorter leg and the hypotenuse.

\begin{align*}2(\overline{AC})=\overline{AB}\end{align*}

Then, substitute the value for \begin{align*}\overline{AC}\end{align*} into the equation and simplify.

\begin{align*}\begin{array}{rcl} 2(\overline{AC}) &=& \overline{AB} \\ 2(4) &=& \overline{AB} \\ 8 &=& \overline{AB} \end{array} \end{align*}

The length of the hypotenuse is 8 feet.

### Review

Find the missing length of the longer leg in each \begin{align*}30^\circ-60^\circ - 90^\circ\end{align*} triangle.

1. Short leg = 3

2. Short leg = 4

3. Short leg = 2

4. Short leg = 8

5. Short leg = 10

Use a calculator to figure out the approximate value of each longer leg. You may round your answer when necessary.

6. \begin{align*}3 \sqrt{3}\end{align*}

7. \begin{align*}4 \sqrt{3}\end{align*}

8. \begin{align*}2 \sqrt{3}\end{align*}

9. \begin{align*}8 \sqrt{3}\end{align*}

10. \begin{align*}10 \sqrt{3} \end{align*}

Use what you have learned to solve each problem.

11. Janie had construction paper cut into and equilateral triangle. She wants to cut it into two smaller congruent triangles. What will be the angle measurement of the triangles that result?

12. Madeleine has poster board in the shape of a square. She wants to cut two congruent triangles out of the poster board without leaving any leftovers. What will be the angle measurements of the triangles that result?

13. A square window has a diagonal of \begin{align*}2 \sqrt{3} \ \text{feet} \end{align*}. What is the length of the shorter of its legs?

14. A square block of cheese is cut into two congruent wedges. If the shortest side of the original block was 9 inches, how long is the diagonal cut?

15. Jerry wants to find the area of an equilateral triangle but only knows that the length of the shorter side is 4 centimeters. What is the height of Jerry’s triangle?

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Color Highlighted Text Notes

### Vocabulary Language: English

30-60-90 Theorem

If a triangle has angle measures of 30, 60, and 90 degrees, then the sides are in the ratio x : x $\sqrt{3}$ : 2x

30-60-90 Triangle

A 30-60-90 triangle is a special right triangle with angles of $30^\circ$, $60^\circ$, and $90^\circ$.

Equilateral Triangle

An equilateral triangle is a triangle in which all three sides are the same length.

Hypotenuse

The hypotenuse of a right triangle is the longest side of the right triangle. It is across from the right angle.

Legs of a Right Triangle

The legs of a right triangle are the two shorter sides of the right triangle. Legs are adjacent to the right angle.

Pythagorean Theorem

The Pythagorean Theorem is a mathematical relationship between the sides of a right triangle, given by $a^2 + b^2 = c^2$, where $a$ and $b$ are legs of the triangle and $c$ is the hypotenuse of the triangle.

The $\sqrt{}$, or square root, sign.