Chapter 8: Right Triangle Trigonometry

The Pythagorean Theorem

Learning Objectives

Identify and employ the Pythagorean Theorem when working with right triangles.
Identify common Pythagorean triples.
Use the Pythagorean Theorem to find the area of isosceles triangles.
Use the Pythagorean Theorem to derive the distance formula on a coordinate grid.

Introduction

The triangle below is a right triangle.

The sides labeled $a$ and $b$ are called the legs of the triangle and they meet at the right angle. The third side, labeled $c$ is called the hypotenuse. The hypotenuse is opposite the right angle. The hypotenuse of a right triangle is also the longest side.

The Pythagorean Theorem states that the length of the hypotenuse squared will equal the sum of the squares of the lengths of the two legs. In the triangle above, the sum of the squares of the legs is $a^2 + b^2$ and the square of the hypotenuse is $c^2$ .

The Pythagorean Theorem: Given a right triangle with legs whose lengths are $a$ and $b$ and a hypotenuse of length $c$ ,

$a^2 + b^2 = c^2$

Be careful when using this theorem—you must make sure that the legs are labeled $a$ and $b$ and the hypotenuse is labeled $c$ to use this equation. A more accurate way to write the Pythagorean Theorem is:

$(\text{leg}_1)^2 + (\text{leg}_2)^2 = \text{hypotenuse}^2$

Example 1

Use the side lengths of the following triangle to test the Pythagorean Theorem.

The legs of the triangle above are $3\;\mathrm{inches}$ and $4\;\mathrm{inches}$ . The hypotenuse is $5\;\mathrm{inches}$ . So, $a = 3 , b = 4$ , and $c = 5$ . We can substitute these values into the formula for the Pythagorean Theorem to verify that the relationship works:

$a^2+b^2 &= c^2\ 3^2+4^2 &= 5^2\ 9+16 &= 25\ 25 &=25$

Since both sides of the equation equal $25$ , the equation is true. Therefore, the Pythagorean Theorem worked on this right triangle.

Proof of the Pythagorean Theorem

There are many ways to prove the Pythagorean Theorem. One of the most straightforward ways is to use similar triangles. Start with a right triangle and construct an altitude from the right angle to the opposite sides. In the figure below, we can see the following relationships:

Proof.

Given: $\triangle WXY$ as shown in the figure below
Prove: $a^{2}+b^{2}=c^{2}$

First we start with a triangle similarity statement:

$\triangle{WXY} \sim \triangle{WZX} \sim \triangle{XZY}$

Now, using similar triangles, we can set up the following proportion:

These are all true by the $AA$ triangle similarity postulate.

$\frac{d}{a} &= \frac{a}{c} \ a^2 &= dc$

and

$\frac{e}{b} &= \frac{b}{c} \ b^2 &= ec$

Putting these equations together by using substitution,

$a^2 + b^2 = dc + ec$

factoring the right hand side,

$a^2 + b^2 = c(d+e)$

but notice $d + e = c$ , so this becomes

$a^2 + b^2 = c(c)$

$a^2 + b^2 = c^2. \blacklozenge$

We have finished proving the Pythagorean Theorem. There are hundreds of other ways to prove the Pythagorean Theorem and one of those alternative proofs is in the exercises for this section.

Making Use of the Pythagorean Theorem

As you know from algebra, if you have one unknown variable in an equation, you can solve to find its value. Therefore, if you know the lengths of two out of three sides in a right triangle, you can use the Pythagorean Theorem to find the length of the missing side, whether it is a leg or a hypotenuse. Be careful to use inverse operations properly and avoid careless mistakes.

Example 2

What is the length of $b$ in the triangle below?

Use the Pythagorean Theorem to find the length of the missing leg, $b$ . Set up the equation $a^2+b^2=c^2$ , letting $a=6$ and $b=10$ . Be sure to simplify the exponents and roots carefully, remember to use inverse operations to solve the equation, and always keep both sides of the equation balanced.

$a^2 + b^2 &= c^2\ 6^2+b^2&=10^2\ 36+b^2&=100\ 36+b^2-36&=100-36\ b^2&=64\ \sqrt{b^2}&=\sqrt{64}\ b&=\pm 8\ b&=8$

In algebra you learned that $\sqrt{x^2}=\pm x$ because, for example, $(5)^2=(-5)^2=25$ . However, in this case (and in much of geometry), we are only interested in the positive solution to $b=\sqrt{64}$ because geometric lengths are positive. So, in example 2, we can disregard the solution $b=-8$ , and our final answer is $b=8\;\mathrm{inches}$ .

Example 3

Find the length of the missing side in the triangle below.

Use the Pythagorean Theorem to set up an equation and solve for the missing side. Let $a = 5$ and $b = 12$ .

$a^2 + b^2 &= c^2\ 5^2+12^2&=c^2\ 25+144&=c^2\ 169&=c^2\ \sqrt{169}&=\sqrt{c^2}\ 13&=c$

So, the length of the missing side is $13\;\mathrm{centimeters}$ .

Using Pythagorean Triples

In example 1, the sides of the triangle were $3 , 4$ , and $5$ . This combination of numbers is referred to as a Pythagorean triple. A Pythagorean triple is three numbers that make the Pythagorean Theorem true and they are integers (whole numbers with no decimal or fraction part). Throughout this chapter, you will use other Pythagorean triples as well. For instance, the triangle in example 2 is proportional to the same ratio of $3:4:5$ . If you divide the lengths of the triangle in example 2 $(6, 8, 10)$ by two, you find the same proportion— $3:4:5$ . Whenever you find a Pythagorean triple, you can apply those ratios with greater factors as well. Finally, take note of the side lengths of the triangle in example 3— $5:12:13$ . This, too, is a Pythagorean triple. You can extrapolate that this ratio, multiplied by greater factors, will also yield numbers that satisfy the Pythagorean Theorem.

There are infinitely many Pythagorean triples, but a few of the most common ones and their multiples are:

Triple	$\times 2$	$\times 3$	$\times 4$
$3-4-5$	$6-8-10$	$9-12-15$	$12-16-20$
$5-12-13$	$10-24-26$	$15-36-39$	$20-48-52$
$7-24-25$	$14-48-50$	$21-72-75$	$28-96-100$
$8-15-17$	$16-30-34$	$24-45-51$	$32-60-68$

Area of an Isosceles Triangle

There are many different applications of the Pythagorean Theorem. One way to use The Pythagorean Theorem is to identify the heights in isosceles triangles so you can calculate the area. The area of a triangle is half of the product of its base and its height (also called altitude). This formula is shown below.

$A = \frac{1} {2} bh$

If you are given the base and the sides of an isosceles triangle, you can use the Pythagorean Theorem to calculate the height. Recall that the height (altitude) of a triangle is the length of a segment from one angle in the triangle perpendicular to the opposite side. In this case we focus on the altitude of isosceles triangles going from the vertex angle to the base.

Example 4

What is the height of the triangle below?

To find the area of this isosceles triangle, you will need to know the height in addition to the base. Draw in the height by connecting the vertex of the triangle with the base at a right angle.

Since the triangle is isosceles, the altitude will bisect the base. That means that it will divide it into two congruent, or equal parts. So, you can identify the length of one half of the base as $4\;\mathrm{centimeters}$ .

If you look at the smaller triangle now inscribed in the original shape, you’ll notice that it is a right triangle with one leg $4$ and hypotenuse $5.$ So, this is a $3:4:5$ triangle. If the leg is $4\;\mathrm{cm}$ and the hypotenuse is $5\;\mathrm{cm}$ , the missing leg must be $3\;\mathrm{cm}$ . So, the height of the isosceles triangle is $3\;\mathrm{cm}$ .

Use this information along with the original measurement of the base to find the area of the entire isosceles triangle.

$A &= \frac{1} {2} bh\ &= \frac{1} {2} (8) (3)\ &= \frac{1} {2} (24)\ &= 12$

The area of the entire isosceles triangle is $12\;\mathrm{cm}^2$ .

The Distance Formula

You have already learned that you can use the Pythagorean Theorem to understand different types of right triangles, find missing lengths, and identify Pythagorean triples. You can also apply the Pythagorean Theorem to a coordinate grid and learn how to use it to find distances between points.

Example 5

Look at the points on the grid below.

Find the length of the segment connecting $(1,5)$ and $(5,2)$ .

The question asks you to identify the length of the segment. Because the segment is not parallel to either axis, it is difficult to measure given the coordinate grid. However, it is possible to think of this segment as the hypotenuse of a right triangle. Draw a vertical line at $x = 1$ and a horizontal line at $y = 2$ and find the point of intersection. This point represents the third vertex in the right triangle.

You can easily count the lengths of the legs of this triangle on the grid. The vertical leg extends from $(1,2)$ to $(1,5)$ , so it is $|5-2|=|3|=3\;\mathrm{units}$ long. The horizontal leg extends from $(1,2)$ to $(5,2)$ , so it is $|5-1|=|4|=4\;\mathrm{units}$ long. Use the Pythagorean Theorem with these values for the lengths of each leg to find the length of the hypotenuse.

$a^2 + b^2 &= c^2\ 3^2+4^2&=c^2\ 9+16&=c^2\ 25&=c^2\ \sqrt{25}&=\sqrt{c^2}\ 5&=c$

The segment connecting $(1,5)$ and $(5,2)$ is $5\;\mathrm{units}$ long.

Mathematicians have simplified this process and created a formula that uses these steps to find the distance between any two points in the coordinate plane. If you use the distance formula, you don’t have to draw the extra lines.

Distance Formula: Give points $(x_1,y_1)$ and $(x_2,y_2)$ , the length of the segment connecting those two points is $D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Example 6

Use the distance formula $D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ to find the distance between the points $(1,5)$ and $(5,2)$ on a coordinate grid.

You already know from example 1 that the distance will be $5\;\mathrm{units}$ , but you can practice using the distance formula to make sure it works. In this formula, substitute $1$ for $x_1 , 5$ for $y_1 , 5$ for $x_2$ , and $2$ for $y_2$ because $(1,5)$ and $(5,2)$ are the two points in question.

$D &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\ &= \sqrt{(5 - 1)^2 + (2 - 5)^2}\ &= \sqrt{(4)^2 + (-3)^2}\ &= \sqrt{16 + 9}\ &= \sqrt{25}\ &= 5$

Now you see that no matter which method you use to solve this problem, the distance between $(1,5)$ and $(5,2)$ on a coordinate grid is $5\;\mathrm{units}$ .

Lesson Summary

In this lesson, we explored how to work with different radical expressions both in theory and in practical situations. Specifically, we have learned:

How to identify and employ the Pythagorean Theorem when working with right triangles.
How to identify common Pythagorean triples.
How to use the Pythagorean Theorem to find the area of isosceles triangles.
How to use the Pythagorean Theorem to derive the distance formula on a coordinate grid.

These skills will help you solve many different types of problems. Always be on the lookout for new and interesting ways to apply the Pythagorean Theorem to mathematical situations.

Points to Consider

Now that you have learned the Pythagorean Theorem, there are countless ways to apply it. Could you use the Pythagorean Theorem to prove that a triangle contained a right angle if you did not have an accurate diagram?

Review Questions

What is the distance between $(-5,-5)$ and $(-2,-1)$ ?
Do the numbers $12$ , $16$ , and $20$ make a Pythagorean triple?
What is the length of $p$ in the triangle below?
Do the numbers $13 , 26$ , and $35$ make a Pythagorean triple?
What is the distance between $(1,9)$ and $(9,4)$ ?
What is the length of $m$ in the triangle below?
What is the distance between $(-3,7)$ and $(6,5)$ ?
What is the area of $\triangle TRN$ below?
What is the area of the triangle below?
What is the area of the triangle below?
An alternative proof of the Pythagorean Theorem uses the area of a square. The diagram below shows a square with side lengths $a + b$ , and an inner square with side lengths $c$ . Use the diagram below to prove $a^2 + b^2 = c^2.$
Hint: Find the area of the inner square in two ways: once directly, and once by finding the area of the larger square and subtracting the area of each triangle.

Review Answers

$5$
yes
$17\;\mathrm{inches}$
no
$\sqrt{89}$
$15\;\mathrm{inches}$
$\sqrt{85}$
$300\;\mathrm{square millimeters}$
$240\;\mathrm{square feet}$
$60\;\mathrm{square yards}$
Proof. The plan is, we will find the area of the green square in two ways. Since those two areas must be equal, we can set those areas equal to each other.
For the inner square (in green), we can directly compute the area: $\text{Area of inner square} = c^2$ .

Now, the area of the large, outer square is $(a+b)^2$ . Don’t forget to multiply this binomial carefully!

$\text{area} &= (a+b)^2\ &=(a+b)(a+b)\ &=a^2+2ab+b^2$

The area of each small right triangle (in yellow) is

$\text{area} = \frac{1}{2} ab$ .

Since there are four of those right triangles, we have the combined area

$4\left(\frac {1}{2}ab \right) = 2ab$

Finally, subtract the area of the four yellow triangles from the area of the larger square, and we are left with

$\text{large square}-\text{four triangles} & = \text{area of inner square}\ a^2+ 2ab + b^2 -2ab & = a^2 + b^2$

Putting together the two different ways for finding the area of the inner square, we have $a^2 + b^2 = c^2.$

Converse of the Pythagorean Theorem

Learning Objectives

Understand the converse of the Pythagorean Theorem.
Identify acute triangles from side measures.
Identify obtuse triangles from side measures.
Classify triangles in a number of different ways.

Converse of the Pythagorean Theorem

In the last lesson, you learned about the Pythagorean Theorem and how it can be used. As you recall, it states that the sum of the squares of the legs of any right triangle will equal the square of the hypotenuse. If the lengths of the legs are labeled $a$ and $b$ , and the hypotenuse is $c$ , then we get the familiar equation:

$a^2 + b^2 = c^2$

The Converse of the Pythagorean Theorem is also true. That is, if the lengths of three sides of a triangle make the equation $a^2+b^2=c^2$ true, then they represent the sides of a right triangle.

With this converse, you can use the Pythagorean Theorem to prove that a triangle is a right triangle, even if you do not know any of the triangle’s angle measurements.

Example 1

Does the triangle below contain a right angle?

This triangle does not have any right angle marks or measured angles, so you cannot assume you know whether the triangle is acute, right, or obtuse just by looking at it. Take a moment to analyze the side lengths and see how they are related. Two of the sides $(15$ and $17)$ are relatively close in length. The third side $(8)$ is about half the length of the two longer sides.

To see if the triangle might be right, try substituting the side lengths into the Pythagorean Theorem to see if they makes the equation true. The hypotenuse is always the longest side, so $17$ should be substituted for $c$ . The other two values can represent $a$ and $b$ and the order is not important.

$a^2 + b^2 &= c^2\ 8^2 + 15^2 &= 17^2\ 64 + 225 &= 289\ 289 &= 289$

Since both sides of the equation are equal, these values satisfy the Pythagorean Theorem. Therefore, the triangle described in the problem is a right triangle.

In summary, example 1 shows how you can use the converse of the Pythagorean Theorem. The Pythagorean Theorem states that in a right triangle with legs $a$ and $b$ , and hypotenuse $c$ , $a^2 + b^2 = c^2$ . The converse of the Pythagorean Theorem states that if $a^2 + b^2 = c^2$ , then the triangle is a right triangle.

Identifying Acute Triangles

Using the converse of the Pythagorean Theorem, you can identify whether triangles contain a right angle or not. However, if a triangle does not contain a right angle, you can still learn more about the triangle itself by using the formula from Pythagorean Theorem. If the sum of the squares of the two shorter sides of a triangle is greater than the square of the longest side, the triangle is acute (all angles are less than $90^\circ$ ). In symbols, if $a^2+b^2>c^2$ then the triangle is acute.

Identifying the "shorter" and "longest" sides may seem ambiguous if sides have the same length, but in this case any ordering of equal length sides leads to the same result. For example, an equilateral triangle always satisfies $a^2+b^2 > c^2$ and so is acute.

Example 2

Is the triangle below acute or right?

The two shorter sides of the triangle are $8$ and $13$ . The longest side of the triangle is $15$ . First find the sum of the squares of the two shorter legs.

$8^2 + 13^2 &= c^2\ 64+169 &=c^2\ 233&=c^2$

The sum of the squares of the two shorter legs is $233.$ Compare this to the square of the longest side, $15.$

$15^2=225$

The square of the longest side is $225.$ Since $8^2+13^2=233\neq 255=15^2$ , this triangle is not a right triangle. Compare the two values to identify which is greater.

$233 > 225$

The sum of the square of the shorter sides is greater than the square of the longest side. Therefore, this is an acute triangle.

Identifying Obtuse Triangles

As you have probably figured out, you can prove a triangle is obtuse (has one angle larger than $90^\circ$ ) by using a similar method. Find the sum of the squares of the two shorter sides in a triangle. If this value is less than the square of the longest side, the triangle is obtuse. In symbols, if $a^2+b^2<c^2$ , then the triangle is obtuse. You can solve this problem in a manner almost identical to example 2 above.

Example 3

Is the triangle below acute or obtuse?

The two shorter sides of the triangle are $5$ and $6.$ The longest side of the triangle is $10.$ First find the sum of the squares of the two shorter legs.

$a^2 + b^2 &= 5^2+6^2\ &= 25+36\ &= 61$

The sum of the squares of the two shorter legs is $61.$ Compare this to the square of the longest side, $10.$

$10^2=100$

The square of the longest side is $100$ . Since $5^2+6^2 \neq 100^2$ , this triangle is not a right triangle. Compare the two values to identify which is greater.

$61 < 100$

Since the sum of the square of the shorter sides is less than the square of the longest side, this is an obtuse triangle.

Triangle Classification

Now that you know the ideas presented in this lesson, you can classify any triangle as right, acute, or obtuse given the length of the three sides. Begin by ordering the sides of the triangle from smallest to largest, and substitute the three side lengths into the equation given by the Pythagorean Theorem using $a \le b < c$ . Be sure to use the longest side for the hypotenuse.

If $a^2 + b^2 = c^2$ , the figure is a right triangle.
If $a^2 + b^2 > c^2$ , the figure is an acute triangle.
If $a^2 + b^2 < c^2$ , the figure is an obtuse triangle.

Example 4

Classify the triangle below as right, acute, or obtuse.

The two shorter sides of the triangle are $9$ and $11$ . The longest side of the triangle is $14$ . First find the sum of the squares of the two shorter legs.

$a^2+b^2 &= 9^2 + 11^2\ &= 81 + 121 \ &= 202$

The sum of the squares of the two shorter legs is $202.$ Compare this to the square of the longest side, $14.$

$14^2=196$

The square of the longest side is $196.$ Therefore, the two values are not equal, $a^2+b^2\neq c^2$ and this triangle is not a right triangle. Compare the two values, $a^2+b^2$ and $c^2$ to identify which is greater.

$202 > 196$

Since the sum of the square of the shorter sides is greater than the square of the longest side, this is an acute triangle.

Example 5

Classify the triangle below as right, acute, or obtuse.

The two shorter sides of the triangle are $16$ and $30$ . The longest side of the triangle is $34$ . First find the sum of the squares of the two shorter legs.

$a^2+b^2 &= 16^2 + 30^2\ &= 256+900\ &=1156$

The sum of the squares of the two legs is $1,156$ . Compare this to the square of the longest side, $34$ .

$c^2=34^2=1156$

The square of the longest side is $1,156.$ Since these two values are equal, $a^2+b^2=c^2$ , and this is a right triangle.

Lesson Summary

In this lesson, we explored how to work with different radical expressions both in theory and in practical situations. Specifically, we have learned:

How to use the converse of the Pythagorean Theorem to prove a triangle is right.
How to identify acute triangles from side measures.
How to identify obtuse triangles from side measures.
How to classify triangles in a number of different ways.

These skills will help you solve many different types of problems. Always be on the lookout for new and interesting ways to apply the Pythagorean Theorem and its converse to mathematical situations.

Points to Consider

Use the Pythagorean Theorem to explore relationships in common right triangles. Do you find that the sides are proportionate?

Review Questions

Solve each problem.

For exercises 1-8, classify the following triangle as acute, obtuse, or right based on the given side lengths. Note, the figure is not to scale.

$a = 9\;\mathrm{in},\ b = 12\;\mathrm{in},\ c = 15\;\mathrm{in}$
$a = 7\;\mathrm{cm},\ b = 7\;\mathrm{cm}, c = 8\;\mathrm{cm}$
$a = 4\;\mathrm{m},\ b = 8\;\mathrm{m},\ c = 10\;\mathrm{m}$
$a = 10\;\mathrm{ft},\ b = 22\;\mathrm{ft}, c = 23\;\mathrm{ft}$
$a = 21\;\mathrm{cm},\ b = 28\;\mathrm{cm},\ c = 35\;\mathrm{cm}$
$a = 10\;\mathrm{ft},\ b = 12\;\mathrm{ft},\ c = 14\;\mathrm{ft}$
$a = 15\;\mathrm{m},\ b = 18\;\mathrm{m},\ c = 30\;\mathrm{m}$
$a = 5\;\mathrm{in},\ b = \sqrt{75}\;\mathrm{in},\ c = 110\;\mathrm{in}$
In the triangle below, which sides should you use for the legs (usually called sides $a$ , and $b$ ) and the hypotenuse (usually called side $c$ ), in the Pythagorean theorem? How do you know?
1. $m\angle A=$
2. $m\angle B=$

Review Answers

Right
Acute
Obtuse
Acute
Right
Acute
Obtuse
Obtuse
The side with length $\sqrt 13$ should be the hypotenuse since it is the longest side. The order of the legs does not matter
$m\angle A=45^\circ , m\angle B=90^\circ$

Using Similar Right Triangles

Learning Objectives

Identify similar triangles inscribed in a larger triangle.
Evaluate the geometric mean of various objects.
Identify the length of an altitude using the geometric mean of a separated hypotenuse.
Identify the length of a leg using the geometric mean of a separated hypotenuse.

Introduction

In this lesson, you will study figures inscribed, or drawn within, existing triangles. One of the most important types of lines drawn within a right triangle is called an altitude. Recall that the altitude of a triangle is the perpendicular distance from one vertex to the opposite side. By definition each leg of a right triangle is an altitude. We can find one more altitude in a right triangle by adding an auxiliary line segment that connects the vertex of the right angle with the hypotenuse, forming a new right angle.

You may recall this is the figure that we used to prove the Pythagorean Theorem. In right triangle $ABC$ above, the segment $\overline {CD}$ is an altitude. It begins at angle $C$ , which is a right angle, and it is perpendicular to the hypotenuse $\overline {AB}$ . In the resulting figure, we have three right triangles, and all of them are similar.

Inscribed Similar Triangles

You may recall that if two objects are similar, corresponding angles are congruent and their sides are proportional in length. In other words, similar figures are the same shape, but different sizes. To prove that two triangles are similar, it is sufficient to prove that all angle measures are congruent (note, this is NOT true for other polygons. For example, both squares and “long” rectangles have all $90^\circ$ angles, but they are not similar). Use logic, and the information presented above to complete Example 1.

Example 1

Justify the statement that $\triangle {TQR} \sim \triangle {TSQ} \sim \triangle {QSR}$ .

In the figure above, the big triangle $\triangle {TQR}$ is a right triangle with right angle $\angle {Q}$ and $m \angle {R} = 30^\circ$ and $m \angle {T} = 60^\circ$ . So, if $\triangle {TQR} , \triangle {TSQ}$ , and $\triangle {QSR}$ are similar, they will all have angles of $30^\circ , 60^\circ$ , and $90^\circ$ .

First look at $\triangle {TSQ}$ . $m \angle {QST}=90^\circ$ , and $m \angle {T} = 60^\circ$ . Since the sum of the three angles in a triangle always equals $180^\circ$ , the missing angle, $\angle {TQS}$ , must measure $30^\circ$ , since $30 + 60 + 90 = 180$ . Lining up the congruent angles, we can write $\triangle {TQR} \sim \triangle {TSQ}$ .

Now look at $\triangle{QRS} . \angle{QSR}$ has a measure of $90^\circ$ , and $m\angle{R}=30^\circ$ . Since the sum of the three angles in a triangle always equals $180^\circ$ , the missing angle, $\angle {RQS}$ , must measure $60^\circ$ , since $30 + 60 + 90 = 180.$ Now, since the triangles have congruent corresponding angles, $\triangle {QSR}$ and $\triangle {TQR}$ are similar.

Thus, $\triangle {TQR} \sim \triangle {TSQ} \sim \triangle {QSR}$ . Their angles are congruent and their sides are proportional.

Note that you must be very careful to match up corresponding angles when writing triangle similarity statements. Here we should write $\triangle {TQR} \sim \triangle {TSQ} \sim \triangle {QSR}$ . This example is challenging because the triangles are overlapping.

Geometric Means

When someone asks you to find the average of two numbers, you probably think of the arithmetic mean (average). Chances are good you’ve worked with arithmetic means for many years, but the concept of a geometric mean may be new. An arithmetic mean is found by dividing the sum of a set of numbers by the number of items in the set. Arithmetic means are used to calculate overall grades and many other applications. The big idea behind the arithmetic mean is to find a “measure of center” for a group of numbers.

A geometric mean applies the same principles, but relates specifically to size, length, or measure. For example, you may have two line segments as shown below. Instead of adding and dividing, you find a geometric mean by multiplying the two numbers, then finding the square root of the product.

To find the geometric mean of these two segments, multiply the lengths and find the square root of the product.

$\text{mean}&=\sqrt{8 \cdot 2}\ &=\sqrt{16} \ &=4$

So, the geometric mean of the two segments would be a line segment that is $4\;\mathrm{cm}$ in length. Use these concepts and strategies to complete example 2.

Example 2

In $\triangle{BCD}$ below, what is the geometric mean of $BC$ and $CD$ ?

When finding a geometric mean, you first find the product of the items involved. In this case, segment $BC$ is $12\;\mathrm{inches}$ and segment $CD$ is $3\;\mathrm{inches}$ . Then find the square root of this product.

$\text{mean}&=\sqrt{12 \cdot 3}\ &=\sqrt{36} \ &=6$

So, the geometric mean of $BC$ and $CD$ in $\triangle{BCD}$ is $6\;\mathrm{inches}$ .

Altitude as Geometric Mean

In a right triangle, the length of the altitude from the right angle to the hypotenuse is the geometric mean of the lengths of the two segments of the hypotenuse. In the diagram below we can use $\triangle{BDC} \sim \triangle{CDA}$ to create the proportion $\frac{d}{f} = \frac{f}{e}$ . Solving for $f , f = \sqrt{d \cdot e}$ .

You can use this relationship to find the length of the altitude if you know the length of the two segments of the divided hypotenuse.

Example 3

What is the length of the altitude $\overline {AD}$ in the triangle below?

To find the altitude of this triangle, find the geometric mean of the two segments of the hypotenuse. In this case, you need to find the geometric mean of $9$ and $3$ . To find the geometric mean, find the product of the two numbers and then take its square root.

$\text{mean} &= \sqrt{9 \cdot 3}\ &=\sqrt{27}\ &=3\sqrt{3}$

So $AD=3\sqrt{3}\;\mathrm{feet}$ , or approximately $3 (1.732) = 5.2\;\mathrm{feet}$ .

Example 4

What is the length of the altitude in the triangle below?

The altitude of this triangle is $\overline{AD}$ . Remember the altitude does not always go “down”! To find $AD$ , find the geometric mean of the two segments of the hypotenuse. Make sure that you fill in missing information in the diagram. You know that the whole hypotenuse, $\overline{CB}$ is $20\;\mathrm{inches}$ long and $BD=4\;\mathrm{inches}$ , but you need to know $CD$ , the length of the longer subsection of $\overline{CB}$ , to find the geometric mean. To do this, subtract.

$CD &= CB-DB\ &=20 - 4\ &=16$

So $CD=16\;\mathrm{inches}$ . Write this measurement on the diagram to keep track of your work.

Now find the geometric mean of $16$ and $4$ to identify the length of the altitude.

$AD &= \sqrt{16 \cdot 4}\ &=\sqrt{64}\ &=8$

The altitude of the triangle will measure $8\;\mathrm{inches}$ .

Leg as Geometric Mean

Just as we used similar triangles to create a proportion using the altitude, the lengths of the legs in right triangles can also be found with a geometric mean with respect to the hypotenuse. The length of one leg in a right triangle is the geometric mean of the adjacent segment and the entire hypotenuse. The diagram below shows the relationships.

$a & = \sqrt{d \cdot c} \ b & = \sqrt{e \cdot c}$

You can use this relationship to find the length of the leg if you know the length of the two segments of the divided hypotenuse.

Example 5

What is the length of $x$ in the triangle below?

To find $x$ , the leg of the large right triangle, find the geometric mean of the adjacent segments of the hypotenuse and the entire hypotenuse. In this case, you need to find the geometric mean of $6$ and $12$ . To find the geometric mean, find the product of the two numbers and then take the square root of that product.

$x &= \sqrt{6 \cdot 12}\ &= \sqrt{72}\ &= 6 \sqrt{2}$

So, $x=6 \sqrt{2}\;\mathrm{millimeters}$ or approximately $8.49\;\mathrm{millimeters}$ .

Example 6

If $m = AB$ , what is the value $m$ in the triangle below?

To find $m$ in this triangle, find the geometric mean of the adjacent segment of the hypotenuse and the entire hypotenuse. Make sure that you fill in missing information in the diagram. You know that the two shorter sections of the hypotenuse are $15\;\mathrm{inches}$ and $5\;\mathrm{inches}$ , but you need to know the length of the entire hypotenuse to find the geometric mean. To do this, add.

$AD+DC &= AC\ 5 + 15 &= 20$

So, $AC = 20\;\mathrm{inches}$ . Write this measurement on the diagram to keep track of your work.

Now find the geometric mean of $20$ and $5$ to identify the length of the altitude.

$m &= \sqrt{20 \cdot 5}\ &= \sqrt{100}\ &= 10$

So, $m=10\;\mathrm{inches}$ .

Lesson Summary

In this lesson, we explored how to work with different radicals both in theory and in practical situations. Specifically, we have learned:

How to identify similar triangles inscribed in a larger triangle.
How to evaluate the geometric mean of various objects.
How to identify the length of an altitude using the geometric mean of a separated hypotenuse.
How to identify the length of a leg using the geometric mean of a separated hypotenuse.

These skills will help you solve many different types of problems. Always be on the lookout for new and interesting ways to find relationships between sides and angles in triangles.

Points to Consider

How can you use the Pythagorean Theorem to identify other relationships between sides in triangles?

Review Questions

Which triangles in the diagram below are similar?
What is the geometric mean of two line segments that are $1$ and $4\;\mathrm{inches}$ , respectively?
What is the geometric mean of two line segments that are $3\;\mathrm{cm}$ each?
Which triangles in the diagram below are similar?
What is the length of the altitude, $h$ , in the triangle below?
What is the length of $d$ in the triangle below?
What is the geometric mean of two line segments that are $4\;\mathrm{yards}$ and $8\;\mathrm{yards}$ , respectively?
What is the length of the altitude in the triangle below?

Use the following diagram from exercises 9-11:
$g =$ ____
$h =$ ____
$k =$ ____ (for an extra challenge, find $k$ in two different ways)
What is the length of the altitude in the triangle below?

Review Answers

Triangles $DEF, EGF,$ and $DGE$ are all similar
$2\;\mathrm{inches}$
$3\;\mathrm{cm}$
Triangles $MNO, PNM,$ and $PMO$ are all similar.
$6\;\mathrm{inches}$
$2\sqrt6\;\mathrm{mm}$ , or approximately $4.9\;\mathrm{mm}$
$4\sqrt2 \;\mathrm{yards}$ , or approximately $5.66\;\mathrm{yards}$
$5\sqrt2 \;\mathrm{feet}$ , or approximately $7.07\;\mathrm{feet}$
$g=\sqrt91 \;\mathrm{inches}$ , or approximately $9.54\;\mathrm{inches}$
$h=\sqrt78 \;\mathrm{inches}$ or approximately $8.83\;\mathrm{inches}$
$k=\sqrt42\;\mathrm{inches}$ or approximately $6.48\;\mathrm{inches}$ . One way to find $k$ is with the geometric mean: $k=\sqrt6.7= \sqrt42\;\mathrm{inches}$ . Alternatively, using the answer from 9 and one of the smaller right triangles, $k = \sqrt{(\sqrt{91})^2- (7)^2}=\sqrt{91-49}= \sqrt{42} \;\mathrm{inches}$
.

Special Right Triangles

Learning Objectives

Identify and use the ratios involved with right isosceles triangles.
Identify and use the ratios involved with $30^\circ-60^\circ-90^\circ$ triangles.
Identify and use ratios involved with equilateral triangles.
Employ right triangle ratios when solving real-world problems.

Introduction

What happens when you cut an equilateral triangle in half using an altitude? You get two right triangles. What about a square? If you draw a diagonal across a square you also get two right triangles. These two right triangles are special special right triangles called the $30^\circ-60^\circ-90^\circ$ and the $45^\circ-45^\circ-90^\circ$ right triangles. They have unique properties and if you understand the relationships between the sides and angles in these triangles, you will do well in geometry, trigonometry, and beyond.

Right Isosceles Triangles

The first type of right triangle to examine is isosceles. As you know, isosceles triangles have two sides that are the same length. Additionally, the base angles of an isosceles triangle are congruent as well. An isosceles right triangle will always have base angles that each measure $45^\circ$ and a vertex angle that measures $90^\circ$ .

Don’t forget that the base angles are the angles across from the congruent sides. They don’t have to be on the bottom of the figure.

Because the angles of all $45^\circ-45^\circ-90^\circ$ triangles will, by definition, remain the same, all $45^\circ-45^\circ-90^\circ$ triangles are similar, so their sides will always be proportional. To find the relationship between the sides, use the Pythagorean Theorem.

Example 1

The isosceles right triangle below has legs measuring $1\;\mathrm{centimeter}$ .

Use the Pythagorean Theorem to find the length of the hypotenuse.

Since the legs are $1\;\mathrm{centimeter}$ each, substitute $1$ for both $a$ and $b$ , and solve for $c$ :

$a^2 + b^2 &= c^2\ 1^2+1^2&=c^2\ 1+1&=c^2\ 2&=c^2\ \sqrt{2}&=\sqrt{c^2}\ c&=\sqrt{2}$

In this example $c = \sqrt{2}\;\mathrm{cm}$ .

What if each leg in the example above was $5\;\mathrm{cm}$ ? Then we would have

$a^2 + b^2 &= c^2\ 5^2+5^2&=c^2\ 25+25&=c^2\ 50&=c^2\ \sqrt{50}&=\sqrt{c^2}\ c&=5\sqrt{2}$

If each leg is $5\;\mathrm{cm}$ , then the hypotenuse is $5\sqrt{2}\;\mathrm{cm}$ .

When the length of each leg was $1$ , the hypotenuse was $1\sqrt{2}$ . When the length of each leg was $5$ , the hypotenuse was $5\sqrt{2}$ . Is this a coincidence? No. Recall that the legs of all $45^\circ-45^\circ-90^\circ$ triangles are proportional. The hypotenuse of an isosceles right triangle will always equal the product of the length of one leg and $\sqrt{2}$ . Use this information to solve the problem in example 2.

Example 2

What is the length of the hypotenuse in the triangle below?

Since the length of the hypotenuse is the product of one leg and $\sqrt{2}$ , you can easily calculate this length. One leg is $4\;\mathrm{inches}$ , so the hypotenuse will be $4\sqrt{2}\;\mathrm{inches}$ , or about $5.66\;\mathrm{inches}$ .

Equilateral Triangles

Remember that an equilateral triangle has sides that all have the same length. Equilateral triangles are also equiangular—all angles have the same measure. In an equilateral triangle, all angles measure exactly $60^\circ$ .

Notice what happens when you divide an equilateral triangle in half.

When an equilateral triangle is divided into two equal parts using an altitude, each resulting right triangle is a $30^\circ-60^\circ-90^\circ$ triangle. The hypotenuse of the resulting triangle was the side of the original, and the shorter leg is half of an original side. This is why the hypotenuse is always twice the length of the shorter leg in a $30^\circ-60^\circ-90^\circ$ triangle. You can use this information to solve problems about equilateral triangles.

30º-60º-90º Triangles

Another important type of right triangle has angles measuring $30^\circ$ , $60^\circ$ , and $90^\circ$ . Just as you found a constant ratio between the sides of an isosceles right triangle, you can find constant ratios here as well. Use the Pythagorean Theorem to discover these important relationships.

Example 3

Find the length of the missing leg in the following triangle. Use the Pythagorean Theorem to find your answer.

Just like you did for $45^\circ-45^\circ-90^\circ$ triangles, use the Pythagorean theorem to find the missing side. In this diagram, you are given two measurements: the hypotenuse $(c)$ is $2\;\mathrm{cm}$ and the shorter leg $(a)$ is $1\;\mathrm{cm}$ . Find the length of the missing leg $(b)$ .

$a^2 + b^2 &= c^2\ 1^2+b^2&=2^2 \ 1+b^2&=4 \ b^2&=3 \ b&= \sqrt{3}$

You can leave the answer in radical form as shown, or use your calculator to find the approximate value of $b \approx 1.732\;\mathrm{cm}$ .

On your own, try this again using a hypotenuse of $6\;\mathrm{feet}$ . Recall that since the $30^\circ-60^\circ-90^\circ$ triangle comes from an equilateral triangle, you know that the length of the shorter leg is half the length of the hypotenuse.

Now you should be able to identify the constant ratios in $30^\circ-60^\circ-90^\circ$ triangles. The hypotenuse will always be twice the length of the shorter leg, and the longer leg is always the product of the length of the shorter leg and $\sqrt{3}$ . In ratio form, the sides, in order from shortest to longest are in the ratio $x:x\sqrt{3}:2x$ .

Example 4

What is the length of the missing leg in the triangle below?

Since the length of the longer leg is the product of the shorter leg and $\sqrt{3}$ , you can easily calculate this length. The short leg is $8\;\mathrm{inches}$ , so the longer leg will be $8\sqrt{3}\;\mathrm{inches}$ , or about $13.86\;\mathrm{inches}$ .

Example 5

What is $AC$ below?

To find the length of segment $\overline {AC}$ , identify its relationship to the rest of the triangle. Since it is an altitude, it forms two congruent triangles with angles measuring $30^\circ$ , $60^\circ$ , and $90^\circ$ . So, $AC$ will be the product of $BC$ (the shorter leg) and $\sqrt {3}$ .

$AC &= BC \sqrt{3} \ &=4 \sqrt{3}$

$AC=4\sqrt{3}\;\mathrm{yards}$ , or approximately $6.93\;\mathrm{yards}$ .

Special Right Triangles in the Real World

You can use special right triangles in many real-world contexts. Many real-life applications of geometry rely on special right triangles, so being able to recall and use these ratios is a way to save time when solving problems.

Example 6

The diagram below shows the shadow a flagpole casts at a certain time of day.

If the length of the shadow cast by the flagpole is $13\;\mathrm{m}$ , what is the height of the flagpole and the length of the hypotenuse of the right triangle shown?

The wording in this problem is complicated, but you only need to notice a few things. You can tell in the picture that this triangle has angles of $30^\circ , 60^\circ$ , and $90^\circ$ (This assumes that the flagpole is perpendicular to the ground, but that is a safe assumption). The height of the flagpole is the longer leg in the triangle, so use the special right triangle ratios to find the length of the hypotenuse.

The longer leg is the product of the shorter leg and $\sqrt{3}$ . The length of the shorter leg is given as $13\;\mathrm{meters}$ , so the height of the flagpole is $13\sqrt{3}\;\mathrm{m}$ .

The length of the hypotenuse is the hypotenuse of a $30^\circ-60^\circ-90^\circ$ triangle. It will always be twice the length of the shorter leg, so it will equal $13 \cdot 2$ , or $26\;\mathrm{meters}$ .

Example 7

Antonio built a square patio in his backyard.

He wants to make a water pipe for flowers that goes from one corner to another, diagonally. How long will that pipe be?

The first step in a word problem of this nature is to add important information to the drawing. Because the problem asks you to find the length from one corner to another, you should draw that segment in.

Once you draw the diagonal path, you can see how triangles help answer this question. Because both legs of the triangle have the same measurement $(17\;\mathrm{feet})$ , this is an isosceles right triangle. The angles in an isosceles right triangle are $45^\circ , 45^\circ$ , and $90^\circ$ .

In an isosceles right triangle, the hypotenuse is always equal to the product of the length of one leg and $\sqrt{2}$ . So, the length of Antonio’s water pipe will be the product of $17$ and $\sqrt{2}$ , or $17 \sqrt{2} \approx 17(1.414)\;\mathrm{feet}$ . This value is approximately equal to $24.04\;\mathrm{feet}$ .

Lesson Summary

In this lesson, we explored how to work with different radicals both in theory and in practical situations. Specifically, we have learned:

How to identify and use the ratios involved with right isosceles triangles.
How to identify and use the ratios involved with $30^\circ-60^\circ-90^\circ$ triangles.
How to identify and use ratios involved with equilateral triangles.
How to employ right triangle ratios when solving real-world problems.

These skills will help you solve many different types of problems. Always be on the lookout for new and interesting ways to find relationships between sides and angles in triangles.

Review Questions

Mildred had a piece of scrap wood cut into an equilateral triangle. She wants to cut it into two smaller congruent triangles. What will be the angle measurement of the triangles that result?
Roberto has a square pizza. He wants to cut two congruent triangles out of the pizza without leaving any leftovers. What will be the angle measurements of the triangles that result?
What is the length of the hypotenuse in the triangle below?
What is the length of the hypotenuse in the triangle below?
What is the length of the longer leg in the triangle below?
What is the length of one of the legs in the triangle below?
What is the length of the shorter leg in the triangle below?
A square window has a diagonal of $5\sqrt2\;\mathrm{feet}$ . What is the length of one of its sides?
A square block of foam is cut into two congruent wedges. If a side of the original block was $3\;\mathrm{feet}$ , how long is the diagonal cut?
They wants to find the area of an equilateral triangle but only knows that the length of one side is $6\;\mathrm{inches}$ . What is the height of Thuy’s triangle? What is the area of the triangle?

Review Answers

$30^\circ , 60^\circ$ , and $90^\circ$
$45^\circ , 45^\circ$ , and $90^\circ$
$10$
$11\sqrt2\;\mathrm{cm}$ or approx. $15.56\;\mathrm{cm}$
$6\sqrt3\;\mathrm{miles}$ or approx. $10.39\;\mathrm{miles}$
$3\;\mathrm{mm}$
$14\;\mathrm{feet}$
$5\;\mathrm{feet}$
$3\sqrt2\;\mathrm{feet}$ or approx. $4.24\;\mathrm{feet}$
$3\sqrt3\;\mathrm{inches}$ or approx. $5.2\;\mathrm{in}$ . The area is $9\sqrt3\approx 15.59\; \mathrm{inches}^2$

Tangent Ratio

Learning Objectives

Identify the different parts of right triangles.
Identify and use the tangent ratio in a right triangle.
Identify complementary angles in right triangles.
Understand tangent ratios in special right triangles.

Introduction

Now that you are familiar with right triangles, the ratios that relate the sides, as well as other important applications, it is time to learn about trigonometric ratios. Trigonometric ratios show the relationship between the sides of a triangle and the angles inside of it. This lesson focuses on the tangent ratio.

Parts of a Triangle

In trigonometry, there are a number of different labels attributed to different sides of a right triangle. They are usually in reference to a specific angle. The hypotenuse of a triangle is always the same, but the terms adjacent and opposite depend on which angle you are referencing. A side adjacent to an angle is the leg of the triangle that helps form the angle. A side opposite to an angle is the leg of the triangle that does not help form the angle.

In the triangle shown above, segment $\overline{AB}$ is adjacent to $\angle{B}$ , and segment $\overline{AC}$ is opposite to $\angle{B}$ . Similarly, $\overline{AC}$ is adjacent to $\angle{x}$ , and $\overline{AB}$ is opposite $\angle{C}$ . The hypotenuse is always $\overline{BC}$ .

Example 1

Examine the triangle in the diagram below.

Identify which leg is adjacent to $\angle{R}$ , opposite to $\angle{R}$ , and the hypotenuse.

The first part of the question asks you to identify the leg adjacent to $\angle{R}$ . Since an adjacent leg is the one that helps to form the angle and is not the hypotenuse, it must be $\overline{QR}$ . The next part of the question asks you to identify the leg opposite $\angle{R}$ . Since an opposite leg is the leg that does not help to form the angle, it must be $\overline{QS}$ . The hypotenuse is always opposite the right angle, so in this triangle the hypotenuse is segment $\overline{RS}$ .

The Tangent Ratio

The first ratio to examine when studying right triangles is the tangent. The tangent of an angle is the ratio of the length of the opposite side to the length of the adjacent side. The hypotenuse is not involved in the tangent at all. Be sure when you find a tangent that you find the opposite and adjacent sides relative to the angle in question.

For an acute angle measuring $x$ , we define $\tan x = \frac{\text{opposite}}{\text{adjacent}}$ .

Example 2

What are the tangents of $\angle{X}$ and $\angle{Y}$ in the triangle below?

To find these ratios, first identify the sides opposite and adjacent to each angle.

$\tan \angle{X} & = \frac{\text{opposite}} {\text{adjacent}} = \frac{5} {12} \approx{0.417} \\ \tan \angle{Y} & = \frac{\text{opposite}} {\text{adjacent}} = \frac{12} {5} = 2.4$

So, the tangent of $\angle{X}$ is about $0.417$ and the tangent of $\angle{Y}$ is $2.4$ .

It is common to write $\tan X$ instead of $\tan \angle X$ . In this text we will use both notations.

Complementary Angles in Right Triangles

Recall that in all triangles, the sum of the measures of all angles must be $180^\circ$ . Since a right angle has a measure of $90^\circ$ , the remaining two angles in a right triangle must be complementary. Complementary angles have a sum of $90^\circ$ . This means that if you know the measure of one of the smaller angles in a right triangle, you can easily find the measure of the other. Subtract the known angle from $90^\circ$ and you’ll have the measure of the other angle.

Example 3

What is the measure of $\angle{N}$ in the triangle below?

To find $m\angle{N}$ , you can subtract the measure of $\angle{N}$ from $90^\circ$ .

$m\angle{N} + m\angle{O} &= 90 \ m\angle{N} &= 90 - m\angle{O}\ m\angle{N} &= 90 - 27\ m\angle{N} &= 63$

So, the measure of $\angle{N}$ is $63^\circ$ since $\angle{N}$ and $\angle{O}$ are complementary.

Tangents of Special Right Triangles

It may help you to learn some of the most common values for tangent ratios. The table below shows you values for angles in special right triangles.

	$30^\circ$	$45^\circ$	$60^\circ$
Tangent	$\frac{1}{\sqrt{3}} \approx 0.577$	$\frac{1}{1} = 1$	$\frac{\sqrt{3}}{1}\approx 1.732$

Notice that you can derive these ratios from the $30^\circ-60^\circ-90^\circ$ special right triangle. You can use these ratios to identify angles in a triangle. Work backwards from the ratio. If the ratio equals one of these values, you can identify the measurement of the angle.

Example 4

What is $m \angle{J}$ in the triangle below?

Find the tangent of $\angle{J}$ and compare it to the values in the table above.

$\tan{J} &= \frac{\text{opposite}} {\text{adjacent}}\\ &= \frac{5} {5}\\ &= 1$

So, the tangent of $\angle{J}$ is $1$ . If you look in the table, you can see that an angle that measures $45^\circ$ has a tangent of $1$ . So, $m \angle{J}=45^\circ$ .

Example 5

What is $m \angle{Z}$ in the triangle below?

Find the tangent of $\angle{Z}$ and compare it to the values in the table above.

$\tan{Z} &= \frac{\text{opposite}} {\text{adjacent}}\\ &= \frac{5.2} {3}\\ &= 1.7\bar{3}$

So, the tangent of $\angle{Z}$ is about $1.73$ . If you look in the table, you can see that an angle that measures $60^\circ$ has a tangent of $1.732$ . So, $m\angle{z} \approx 60^\circ$ .

Notice in this example that $\triangle{XYZ}$ is a $30^\circ-60^\circ-90^\circ$ triangle. You can use this fact to see that $XY = 5.2 \approx 3\sqrt{3}$ .

Lesson Summary

In this lesson, we explored how to work with different radical expressions both in theory and in practical situations. Specifically, we have learned:

How to identify the different parts of right triangles.
How to identify and use the tangent ratio in a right triangle.
How to identify complementary angles in right triangles.
How to understand tangent ratios in special right triangles.

These skills will help you solve many different types of problems. Always be on the lookout for new and interesting ways to find relationships between sides and angles in triangles.

Review Qusetions

Use the following diagram for exercises 1-5.

How long is the side opposite angle $G?$
How long is the side adjacent to angle $G?$
How long is the hypotenuse?
What is the tangent of $\angle G?$
What is the tangent of $\angle H?$
What is the measure of $\angle C$ in the diagram below?
What is the measure of $\angle H$ in the diagram below?

Use the following diagram for exercises 8-9.
What is the tangent of $\angle R?$
What is the tangent of $\angle S?$
What is the measure of $\angle E$ in the triangle below?

Review Answers

$8\;\mathrm{mm}$
$6\;\mathrm{mm}$
$10\;\mathrm{mm}$
$\frac{8} {6} = 1.\bar{3}$
$\frac{6} {8} = \frac{3} {4} = 0.75$
$32^\circ$
$45^\circ$
$\frac{7} {24} = 0.292$
$\frac{24} {7} = 3.43$
$72^\circ$

Sine and Cosine Ratios

Learning Objectives

Review the different parts of right triangles.
Identify and use the sine ratio in a right triangle.
Identify and use the cosine ratio in a right triangle.
Understand sine and cosine ratios in special right triangles.

Introduction

Now that you have some experience with tangent ratios in right triangles, there are two other basic types of trigonometric ratios to explore. The sine and cosine ratios relate opposite and adjacent sides of a triangle to the hypotenuse. Using these three ratios and a calculator or a table of trigonometric ratios you can solve a wide variety of problems!

Review: Parts of a Triangle

The sine and cosine ratios relate opposite and adjacent sides to the hypotenuse. You already learned these terms in the previous lesson, but they are important to review and commit to memory. The hypotenuse of a triangle is always opposite the right angle, but the terms adjacent and opposite depend on which angle you are referencing. A side adjacent to an angle is the leg of the triangle that helps form the angle. A side opposite to an angle is the leg of the triangle that does not help form the angle.

Example 1

Examine the triangle in the diagram below.

Identify which leg is adjacent to angle $N$ , which leg is opposite to angle $N$ , and which segment is the hypotenuse.

The first part of the question asks you to identify the leg adjacent to $\angle{N}$ . Since an adjacent leg is the one that helps to form the angle and is not the hypotenuse, it must be $\overline{MN}$ . The next part of the question asks you to identify the leg opposite $\angle{N}$ . Since an opposite leg is the leg that does not help to form the angle, it must be $\overline{LM}$ . The hypotenuse is always opposite the right angle, so in this triangle it is segment $\overline{LN}$ .

The Sine Ratio

Another important trigonometric ratio is sine. A sine ratio must always refer to a particular angle in a right triangle. The sine of an angle is the ratio of the length of the leg opposite the angle to the length of the hypotenuse. Remember that in a ratio, you list the first item on top of the fraction and the second item on the bottom.

So, the ratio of the sine will be

$\sin x = \frac{\text{opposite}} {\text{hypotenuse}}.$

Example 2

What are $\sin A$ and $\sin B$ in the triangle below?

All you have to do to find the solution is build the ratio carefully.

$\sin A & = \frac{\text{opposite}} {\text{hypotenuse}} = \frac{3} {5} = 0.6 \ \sin B & = \frac{\text{opposite}} {\text{hypotenuse}} = \frac{4} {5} = 0.8$

So, $\sin A = 0.6$ and $\sin B = 0.8$ .

The Cosine Ratio

The next ratio to examine is called the cosine. The cosine is the ratio of the adjacent side of an angle to the hypotenuse. Use the same techniques you used to find sines to find cosines.

$\cos(\text{angle}) = \frac{\text{adjacent}} {\text{hypotenuse}}$

Example 3

What are the cosines of $\angle{M}$ and $\angle{N}$ in the triangle below?

To find these ratios, identify the sides adjacent to each angle and the hypotenuse. Remember that an adjacent side is the one that does create the angle and is not the hypotenuse.

$\cos M & = \frac{\text{adjacent}} {\text{hypotenuse}} = \frac{15} {17} \approx 0.88 \ \cos N & = \frac{\text{adjacent}} {\text{hypotenuse}} = \frac{8} {17} \approx 0.47$

So, the cosine of $\angle{M}$ is about $0.88$ and the cosine of $\angle{N}$ is about $0.47$ .

Note that $\triangle{LMN}$ is NOT one of the special right triangles, but it is a right triangle whose sides are a Pythagorean triple.

Sines and Cosines of Special Right Triangles

It may help you to learn some of the most common values for sine and cosine ratios. The table below shows you values for angles in special right triangles.

	$30^\circ$	$45^\circ$	$60^\circ$
Sine	$\frac{1}{2}=0.5$	$\frac{1}{\sqrt{2}}\approx 0.707$	$\frac{\sqrt{3}}{2}\approx 0.866$
Cosine	$\frac{\sqrt{3}}{2}\approx 0.866$	$\frac{1}{\sqrt{2}}\approx 0.707$	$\frac{1}{2}=0.5$

You can use these ratios to identify angles in a triangle. Work backwards from the ratio. If the ratio equals one of these values, you can identify the measurement of the angle.

Example 4

What is the measure of $\angle{C}$ in the triangle below?

Note: Figure is not to scale.

Find the sine of $\angle{C}$ and compare it to the values in the table above.

$\sin{C} &= \frac{\text{opposite}} {\text{hypotenuse}}\\ &= \frac{12} {24}\\ &= 0.5$

So, the sine of $\angle{C}$ is $0.5$ . If you look in the table, you can see that an angle that measures $30^\circ$ has a sine of $0.5$ . So, $m\angle{C}= 30^\circ$ .

Example 5

What is the measure of $\angle{G}$ in the triangle below?

Find the cosine of $\angle{G}$ and compare it to the values in the previous table.

$\cos{G} &= \frac{\text{adjacent}} {\text{hypotenuse}}\\ &= \frac{3} {4.24}\\ &= 0.708$

So, the cosine of $\angle{G}$ is about $0.708$ . If you look in the table, you can see that an angle that measures $45^\circ$ has a cosine of $0.707$ . So, $\angle{G}$ measures about $45^\circ$ . This is a $45^\circ-45^\circ-90^\circ$ right triangle.

Lesson Summary

In this lesson, we explored how to work with different trigonometric ratios both in theory and in practical situations. Specifically, we have learned:

The different parts of right triangles.
How to identify and use the sine ratio in a right triangle.
How to identify and use the cosine ratio in a right triangle.
How to apply sine and cosine ratios in special right triangles.

These skills will help you solve many different types of problems. Always be on the lookout for new and interesting ways to find relationships between sides and angles in triangles.

Points to Consider

Before you begin the next lesson, think about strategies you could use to simplify an equation that contains a trigonometric function.

Note, you can only use the $\sin , \cos$ , and $\tan$ ratios on the acute angles of a right triangle. For now it only makes sense to talk about the $\sin , \cos$ , or $\tan$ ratio of an acute angle. Later in your mathematics studies you will redefine these ratios in a way that you can talk about $\sin , \cos$ , and $\tan$ of acute, obtuse, and even negative angles.

Review Questions

Use the following diagram for exercises 1-3.

What is the sine of $\angle V?$
What is the cosine of $\angle V?$
What is the cosine of $\angle U?$
Use the following diagram for exercises 4-6.
What is the sine of $\angle O?$
What is the cosine of $\angle O?$
What is the sine of $\angle M?$
What is the measure of $\angle H$ in the diagram below?

Use the following diagram for exercises 8-9.
What is the sine of $\angle S?$
What is the cosine of $\angle S?$
What is the measure of $\angle E$ in the triangle below?

Review Answers

$\frac{12} {13} \approx 0.923\;\mathrm{cm}$
$\frac{5} {13} \approx 0.385\;\mathrm{cm}$
$\frac{12} {13} \approx 0.923\;\mathrm{cm}$
$\frac{12} {15} = \frac{4} {5} = 0.8\;\mathrm{inches}$
$\frac{9} {15} = \frac{3} {5} = 0.6\;\mathrm{inches}$
$\frac{9} {15} = \frac{3} {5} = 0.6\;\mathrm{inches}$
$45^\circ$
$\frac{8} {17} \approx 0.471$
$\frac{15} {17} \approx 0.882$
$60^\circ$

Inverse Trigonometric Ratios

Learning Objectives

Identify and use the arctangent ratio in a right triangle.
Identify and use the arcsine ratio in a right triangle.
Identify and use the arccosine ratio in a right triangle.
Understand the general trends of trigonometric ratios.

Introduction

The word inverse is probably familiar to you often in mathematics, after you learn to do an operation, you also learn how to “undo” it. Doing the inverse of an operation is a way to undo the original operation. For example, you may remember that addition and subtraction are considered inverse operations. Multiplication and division are also inverse operations. In algebra you used inverse operations to solve equations and inequalities. You may also remember the term additive inverse, or a number that can be added to the original to yield a sum of $0$ . For example, $5$ and $-5$ are additive inverses because $5 + (-5) = 0$ .

In this lesson you will learn to use the inverse operations of the trigonometric functions you have studied thus far. You can use inverse trigonometric functions to find the measures of angles when you know the lengths of the sides in a right triangle.

Inverse Tangent

When you find the inverse of a trigonometric function, you put the word arc in front of it. So, the inverse of a tangent is called the arctangent (or arctan for short). Think of the arctangent as a tool you can use like any other inverse operation when solving a problem. If tangent tells you the ratio of the lengths of the sides opposite and adjacent to an angle, then arctan tells you the measure of an angle with a given ratio.

Suppose $\tan{X} = 0.65$ . The arctangent can be used to find the measure of $\angle{X}$ on the left side of the equation.

$\arctan{(\tan{X})} & = \arctan{(0.65)} \ m\angle{X} & = \arctan{(0.65)} \approx 33^\circ$

Where did that $33^\circ$ come from? There are two basic ways to find an arctangent. Sometimes you will be given a table of trigonometric values and the angles to which they correspond. In this scenario, find the value that is closest to the one provided, and identify the corresponding angle.

Another, easier way of finding the arctangent is to use a calculator. The arctangent button may be labeled “arctan,” “atan,” or “ $\tan^{-1}$ .” Either way, select this button, and input the value in question. In this case, you would press the arctangent button and enter $0.65$ (or on some calculators, enter $.65$ , then press “arctan”). The output will be the value of measure $\angle{X}$ .

$m\angle{X} & = \arctan{(0.65)}\ m\angle{X} & \approx 33$

$m\angle{X}$ is about $33^\circ$ .

Example 1

Solve for $m\angle{Y}$ if $\tan{Y} = 0.384$

You can use the inverse of tangent, arctangent to find this value.

$\arctan{(\tan {Y})} & = \arctan{(0.384)} \ m\angle{Y} & = \arctan{(0.384)}$

Then use your calculator to find the arctangent of $0.384$ .

$m\angle{Y} \approx 21^\circ$

Example 2

What is $m\angle{B}$ in the triangle below?

First identify the proper trigonometric ratio related to $\angle{B}$ that can be found using the sides given. The tangent uses the opposite and adjacent sides, so it will be relevant here.

$\tan{B} &= \frac{\text{opposite}} {\text{adjacent}}\\ &= \frac{8} {5}\\ &= 1.6$

Now use the arctangent to solve for the measure of $\angle{B}$ .

$\arctan{(\tan{B})} & = \arctan{(1.6)} \ m\angle{B} & = \arctan{(1.6)}$

Then use your calculator to find the arctangent of $1.6$ .

$m\angle{B} \approx 58^\circ$

Inverse Sine

Just as you used arctangent as the inverse operation for tangent, you can also use arcsine (shortened as arcsin) as the inverse operation for sine. The same rules apply. You can use it to isolate a variable for an angle measurement, but you must perform the operation on both sides of the equation. When you know the arcsine value, use a table or a calculator to find the measure of the angle.

Example 3

Solve for $m\angle{P}$ if $\sin{P} = 0.891$

You can use the inverse of sine, arcsine to find this value.

$\arcsin{(\sin{P})} & = \arcsin{(0.891)} \ m\angle{P} & = \arcsin{(0.891)}$

Then use your calculator to find the arcsine of $0.891$ .

$m\angle{P} \approx 63^\circ$

Example 4

What is $m \angle{F}$ in the triangle below?

First identify the proper trigonometric ratio related to angle $F$ that can be found using the sides given. The sine uses the opposite side and the hypotenuse, so it will be relevant here.

$\sin{F} & = \frac{\text{opposite}} {\text{adjacent}} \ \sin{F} & = \frac{12} {13} \ \sin{F} & \approx 0.923$

Now use the arcsine to isolate the value of angle $F$ .

$\arcsin{(\sin{F})} & = \arcsin{(0.923)} \ m\angle{F} & = \arcsin{(0.923)}$

Finally, use your calculator to find the arcsine of $0.923$ .

$m\angle{F} \approx 67^\circ$

Inverse Cosine

The last inverse trigonometric ratio is arccosine (often shortened to arccos). The same rules apply for arccosine as apply for all other inverse trigonometric functions. You can use it to isolate a variable for an angle measurement, but you must perform the operation on both sides of the equation. When you know the arccosine value, use a table or a calculator to find the measure of the angle.

Example 5

Solve for $m\angle{Z}$ if $\cos{Z} = 0.31$ .

You can use the inverse of cosine, arccosine, to find this value.

$\arccos{(\cos{Z})} & = \arccos{(0.31)} \ m\angle{Z} & = \arccos{(0.31)}$

Then use your calculator to find the arccosine of $0.31$ .

$m\angle{Z} \approx 72^\circ$

Example 6

What is the measure of $\angle{K}$ in the triangle below?

First identify the proper trigonometric ratio related to $\angle{K}$ that can be found using the sides given. The cosine uses the adjacent side and the hypotenuse, so it will be relevant here.

$\cos{K} &= \frac{\text{adjacent}} {\text{hypotenuse}}\\ &= \frac{9} {11}\\ &= 0.818$

Now use the arccosine to isolate the value of $\angle{K}$ .

$\arccos{(\cos{K})} & = \arccos{(0.818)} \ m\angle{K} & = \arccos{(0.818)}$

Finally use your calculator or a table to find the arccosine of $0.818.$

$m\angle{K} \approx 35^\circ$

General Trends in Trigonometric Ratios

Now that you know how to find the trigonometric ratios as well as their inverses, it is helpful to look at trends in the different values. Remember that each ratio will have a constant value for a specific angle. In any right triangle, the sine of a $30^\circ$ angle will always be $0.5$ —it doesn’t matter how long the sides are. You can use that information to find missing lengths in triangles where you know the angles, or to identify the measure of an angle if you know two of the sides.

Examine the table below for trends. It shows the sine, cosine, and tangent values for eight different angle measures.

	$10^\circ$	$20^\circ$	$30^\circ$	$40^\circ$	$50^\circ$	$60^\circ$	$70^\circ$	$80^\circ$
Sine	$0.174$	$0.342$	$0.5$	$0.643$	$0.766$	$0.866$	$0.940$	$0.985$
Cosine	$0.985$	$0.940$	$0.866$	$0.766$	$0.643$	$0.5$	$0.342$	$0.174$
Tangent	$0.176$	$0.364$	$0.577$	$0.839$	$1.192$	$1.732$	$2.747$	$5.671$

Example 7

Using the table above, which value would you expect to be greater: the sine of $25^\circ$ or the cosine of $25^\circ$ ?

You can use the information in the table to solve this problem. The sine of $20^\circ$ is $0.342$ and the sine of $30^\circ$ is $0.5$ . So, the sine of $25^\circ$ will be between the values $0.342$ and $0.5$ . The cosine of $20^\circ$ is $0.940$ and the cosine of $30^\circ$ is $0.866.$ So, the cosine of $25^\circ$ will be between the values of $0.866$ and $0.940.$ Since the range for the cosine is greater, than the range for the sine, it can be assumed that the cosine of $25^\circ$ will be greater than the sine of $25^\circ.$

Notice that as the angle measures approach $90^\circ$ , $\sin$ approaches $1$ . Similarly, as the value of the angles approach $90^\circ$ , the $\cos$ approaches $0$ . In other words, as the $\sin$ gets greater, the $\cos$ gets smaller for the angles in this table.

The tangent, on the other hand, increases rapidly from a small value to a large value (infinity, in fact) as the angle approaches $90^\circ$ .

Lesson Summary

In this lesson, we explored how to work with different radicals both in theory and in practical situations. Specifically, we have learned:

How to identify and use the arctangent ratio in a right triangle.
How to identify and use the arcsine ratio in a right triangle.
How to identify and use the arccosine ratio in a right triangle.
How to understand the general trends of trigonometric ratios.

These skills will help you solve many different types of problems. Always be on the lookout for new and interesting ways to find relationships between sides and angles in triangles.

Points to Consider

To this point, all of the trigonometric ratios you have studied have dealt exclusively with right triangles. Can you think of a way to use trigonometry on triangles that are acute or obtuse?

Review Questions

Solve for $m\angle{G}.$
$\cos G = 0.53$
Solve for $m\angle{V}.$
$\tan V = 2.25$
What is the measure of $\angle{B}$ in the triangle below?
Solve for $m\angle{M}.$
$\sin M = 0.978$
What is the measure of $\angle{F}$ in the triangle below?
Solve for $m\angle{L}.$
$\tan L = 1.04$
Solve for $m\angle{D}.$
$\cos D = 0.07$
What is the measure of $\angle{M}$ in the triangle below?
What is the measure of $\angle{Q}$ in the triangle below?
What is the measure of $\angle{Z}$ in the triangle below?

Review Answers

$58^\circ$
$66^\circ$
$60^\circ$
$78^\circ$
$70^\circ$
$46^\circ$
$86^\circ$
$89^\circ$
$67^\circ$
$5.7^\circ$

Acute and Obtuse Triangles

Learning Objectives

Identify and use the Law of Sines.
Identify and use the Law of Cosines.

Introduction

Trigonometry is most commonly learned on right triangles, but the ratios can have uses for other types of triangles, too. This lesson focuses on how you can apply sine and cosine ratios to angles in acute or obtuse triangles. Remember that in an acute triangle, all angles measure less than $90^\circ$ . In an obtuse triangle, there will be one angle that has a measure that is greater than $90^\circ$ .

The Law of Sines

The Law of Sines states that in any triangle, the ratio of the length of a side to the sine of the angle opposite it will be constant. That is, the ratio is the same for all three angles and their opposite sides. Thus, if you find the ratio, you can use it to find missing angle measure and side lengths.

$\frac{a} {\sin{A}} = \frac{b} {\sin{B}} = \frac{c} {\sin{C}}$

Note the convention that $A$ denotes $\angle{A}$ and $a$ is the length of the side opposite $\angle{A}$ .

Example 1

Examine the triangle in the following diagram.

What is the length of the side labeled j?

You can use the law of sines to solve this problem. Because you have one side and the angle opposite, you can find the constant that applies to the entire triangle. This ratio will be equal to the proportion of side $j$ and $\angle{J}$ . You can use your calculator to find the value of the sines.

$\frac{h} {\sin{H}} &= \frac{j} {\sin{J}}\ \frac{6} {\sin{38}} &= \frac{7} {\sin{70}} \ \frac{6} {0.616} &= \frac{j} {0.940} \ 9.74 &= \frac{j} {0.940} \ 9.2 &\approx j$

So, using the law of sines, the length of $j$ is about $9.2\;\mathrm{meters}$ .

Example 2

Examine the triangle in the following diagram.

What is the measure of $\angle{S}$ ?

You can use the law of sines to solve this problem. Because you have one side and the angle opposite, you can find the constant that applies to the entire triangle. This ratio will be equal to the proportion of side $r$ and angle $R$ . You can use your calculator to find the value of the sines.

$\frac{r} {\sin{R}} &= \frac{s} {\sin{S}} \ \frac{5} {\sin{53}} &= \frac{5.26} {\sin{S}} \ \frac{5} {0.788} &= \frac{5.26} {\sin{S}} \ 6.345 &= \frac{5.26} {\sin{S}} \ (6.345) \cdot \sin{S} &= 5.26 \ \sin{S} &= \frac{5.26}{6.345} \ \sin{S} &= 0.829 \ \arcsin{(\sin{S})} &= \arcsin{0.829} \ m\angle{S} &\approx 56^\circ$

So, using the law of sines, the angle labeled $S$ must measure about $56^\circ$ .

The Law of Cosines

There is another law that works on acute and obtuse triangles in addition to right triangles. The Law of Cosines uses the cosine ratio to identify either lengths of sides or missing angles. To use the law of cosines, you must have either the measures of all three sides, or the measure of two sides and the measure of the included angle.

$c^2 = a^2 + b^2 - 2ab(\cos{C})$

It doesn’t matter how you assign the variables to the three sides of the triangle, but the angle $C$ must be opposite side $c$ .

Example 3

Examine the triangle in the following diagram.

What is the measure of side $\overline{EF}$ ?

Use the Law of Cosines to find $EF$ . Since $\overline{EF}$ is opposite $\angle{D}$ , we will call the length of $\overline{EF}$ by the letter $d$ .

$d^2 &= e^2 + f^2 -2ab (\cos{D}) \ d^2 &= (6)^2 + (7)^2 -2(6)(7)(\cos{60}) \\ d^2 &= 36 + 49 - 84(\cos{60}) \ d^2 &= 85 - 84(0.5) \ d^2 &= 85 - 42 \ d^2 &= 43 \ d &= \sqrt{43} \ d&\approx 6.56$

So, $EF$ is about $6.56\;\mathrm{inches}$ .

Example 4

Examine the triangle in the following diagram.

What is the measure of $\angle{X}?$

Use the Law of Cosines to find the measure of $\angle{X}$ .

$x^2 &= y^2 + z^2 - 2yz(\cos{X}) \\ (5.39)^2 &= (5)^2 + (7.6)^2 - 2(7.6)(5)(\cos{X}) \ 29.05 &= 25 + 57.76 - 76(\cos{X}) \ 29.05 &= 82.76 - 76(\cos{X}) \ -53.71 &= -76(\cos{X}) \ 0.707 &= (\cos{X}) \ \arccos{(0.707)} &= \arccos{(cos{X})} \ 45^\circ &\approx m\angle{X}$

So, $m\angle{X}$ is about $45^\circ$ .

Lesson Summary

In this lesson, we explored how to work with different radical expressions both in theory and in practical situations. Specifically, we have learned:

how to identify and use the law of sines.
how to identify and use the law of cosines.

These skills will help you solve many different types of problems. Always be on the lookout for new and interesting ways to find relationships between sides and angles in triangles.

Review Questions

Exercises 1 and 2 use the triangle in the following diagram.

What is the length of side $BC?$
What is $m\angle{C}?$
Examine the triangle in the following diagram.

What is the measure of $\angle{F}?$
Examine the triangle in the following diagram.

What is the measure of $\angle{I}?$
Examine the triangle in the following diagram.

What is the measure of side $KL?$
Examine the triangle in the following diagram.

What is the measure of $\angle{O}?$

Use the triangle in the following diagram for exercises 7 and 8.
What is the measure of $\angle{P}?$
What is the measure of $\angle{Q}?$
Examine the triangle in the following diagram.

What is the measure of $\angle{T}?$
Examine the triangle in the following diagram.

What is the measure of $\angle{W}?$

Review Answers

$8\;\mathrm{inches}$
$47.6^\circ$
$48^\circ$
$83^\circ$
$16.5\;\mathrm{inches}$
$22.3^\circ$
$40^\circ$
$48.6^\circ$
$35^\circ$
$78^\circ$