10.8 Right Triangle Trigonometry

Pythagorean Theorem

The Pythagorean Theorem is used to find unknown sides of right triangles. The theorem states that the sum of the squares of the two legs of a right triangle equals the square of the hypotenuse (the longest side of the right triangle).

FORMULA

The Pythagorean Theorem states

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

where $a$ and $b$ are two sides (legs) of a right triangle and $c$ is the hypotenuse, as shown in Figure 10.138.

Figure 10.138 Pythagorean Right Triangle

For example, given that side $a=6,$ and side $b=8,$ we can find the measure of side $c$ using the Pythagorean Theorem. Thus,

Example 10.64

Using the Pythagorean Theorem

Find the length of the missing side of the triangle (Figure 10.139).

Figure 10.139

Solution

Using the Pythagorean Theorem, we have

$$\begin{array}{ccc}\hfill {(6)}^2+b^2& \hfill =\hfill & {(14)}^2\hfill \\ \hfill 36+b^2& \hfill =\hfill & 196\hfill \\ \hfill b^2& \hfill =\hfill & 196-36\hfill \\ \hfill b^2& \hfill =\hfill & 160\hfill \\ \hfill b& \hfill =\hfill & \pm \sqrt{160}\hfill \\ \hfill & \hfill =\hfill & 4\sqrt{10}=12.65\hfill \end{array}$$

When we take the square root of a number, the answer is usually both the positive and negative root. However, lengths cannot be negative, which is why we only consider the positive root.

Your Turn 10.64

1.

Use the Pythagorean Theorem to find the missing side of the right triangle shown.

Distance

The applications of the Pythagorean Theorem are countless, but one especially useful application is that of distance. In fact, the distance formula stems directly from the theorem. It works like this:

In Figure 10.140, the problem is to find the distance between the points $(-3,-1)$ and $(3,2).$ We call the length from point $(-3,-1)$ to point $(3,-1)$ side $a$, and the length from point $(3,-1)$ to point $(3,2)$ side $b$. To find side $c$, we use the distance formula and we will explain it relative to the Pythagorean Theorem. The distance formula is $d=\sqrt{{(x_2-x_1)}^2+{\left(y_2-y_1\right)}^2},$ such that $(x_2-x_1)$ is a substitute for $a$ in the Pythagorean Theorem and is equal to $3-(-3)=6;$ and $(y_2-y_1)$ is a substitute for $b$ in the Pythagorean Theorem and is equal to $2-(-1)=3.$ When we plug in these numbers to the distance formula, we have

Thus, $d=c$, the hypotenuse, in the Pythagorean Theorem.

Figure 10.140 Distance

Example 10.65

Calculating Distance Using the Distance Formula

You live on the corner of First Street and Maple Avenue, and work at Star Enterprises on Tenth Street and Elm Drive (Figure 10.141). You want to calculate how far you walk to work every day and how it compares to the actual distance (as the crow flies). Each block measures 200 ft by 200 ft.

Figure 10.141

Solution

You travel 7 blocks south and 9 blocks west. If each block measures 200 ft by 200 ft, then $9(200)+7(200)=\mathrm{1,800}\text{ft}+\mathrm{1,400}\text{ft}=\mathrm{3,200}\text{ft}$.

As the crow flies, use the distance formula. We have

$$\begin{array}{ccc}\hfill d& \hfill =\hfill & \sqrt{{(\mathrm{1,800}-0)}^2+{(\mathrm{1,400}-0)}^2}\hfill \\ \hfill & \hfill =\hfill & \sqrt{\mathrm{3,240,000}+\mathrm{1,960,000}}\hfill \\ \hfill & \hfill =\hfill & \sqrt{\mathrm{5,200,000}}\hfill \\ \hfill & \hfill =\hfill & 2280.4\text{ft}\hfill \end{array}$$

Example 10.66

Calculating Distance with the Pythagorean Theorem

The city has specific building codes for wheelchair ramps. Every vertical rise of 1 in requires that the horizontal length be 12 inches. You are constructing a ramp at your business. The plan is to make the ramp 130 inches in horizontal length and the slanted distance will measure approximately 132.4 inches (Figure 10.142). What should the vertical height be?

Figure 10.142

Solution

The Pythagorean Theorem states that the horizontal length of the base of the ramp, side a, is 130 in. The length of c, or the length of the hypotenuse, is 132.4 in. The length of the height of the triangle is side b.

Then, by the Pythagorean Theorem, we have:

$$\begin{array}{ccc}\hfill a^2+b^2& \hfill =\hfill & c^2\hfill \\ \hfill {(130)}^2+b^2& \hfill =\hfill & {(132.4)}^2\hfill \\ \hfill \mathrm{16,900}+b^2& \hfill =\hfill & \mathrm{17,529.76}\hfill \\ \hfill b^2& \hfill =\hfill & \mathrm{17,529.76}-\mathrm{16,900}\hfill \\ \hfill b^2& \hfill =\hfill & 629.8\hfill \\ \hfill b& \hfill =\hfill & \sqrt{629.8}=25\hfill \end{array}$$

If you construct the ramp with a 25 in vertical rise, will it fulfill the building code? If not, what will have to change?

The building code states 12 in of horizontal length for each 1 in of vertical rise. The vertical rise is 25 in, which means that the horizontal length has to be $12(25)=300\text{in}.$ So, no, this will not pass the code. If you must keep the vertical rise at 25 in, what will the other dimensions have to be? Since we need a minimum of 300 in for the horizontal length:

$$\begin{array}{ccc}\hfill {(300)}^2+{(25)}^2& \hfill =\hfill & c^2\hfill \\ \hfill \mathrm{90,625}& \hfill =\hfill & c^2\hfill \\ \hfill \sqrt{\mathrm{90,625}}& \hfill =\hfill & c=301\text{in}\hfill \end{array}$$

The new ramp will look like Figure 10.143.

Figure 10.143

Your Turn 10.66

1.

If 10 in is the maximum possible vertical rise as shown in the figure, how long would the ramp have to be to pass the building code rule of 12 horizontal inches to 1 vertical inch?

${30}^{\circ }\text{-}{60}^{\circ }\text{-}{90}^{\circ }$ Triangles

In geometry, as in all fields of mathematics, there are always special rules for special circumstances. An example is the perfect square rule in algebra. When expanding an expression like ${\left(2x+5y\right)}^2,$ we do not have to expand it the long way:

If we know the perfect square formula, given as

we can skip the middle step and just start writing down the answer. This may seem trivial with problems like ${\left(a+b\right)}^2.$ However, what if you have a problem like ${\left(2\sqrt{3}+3\sqrt[3]{31.8c}\right)}^2?$ That is a different story. Nevertheless, we use the same perfect square formula. The same idea applies in geometry. There are special formulas and procedures to apply in certain types of problems. What is needed is to remember the formula and remember the kind of problems that fit. Sometimes we believe that because a formula is labeled special, we will rarely have use for it. That assumption is incorrect. So, let us identify the ${30}^{\circ }\text{-}{60}^{\circ }\text{-}{90}^{\circ }$ triangle and find out why it is special. See Figure 10.144.

Figure 10.144 The $30^{\circ }\text{-}60^{\circ }\text{-}90^{\circ }$

We see that the shortest side is opposite the smallest angle, and the longest side, the hypotenuse, will always be opposite the right angle. There is a set ratio of one side to another side for the ${30}^{\circ }\text{-}{60}^{\circ }\text{-}{90}^{\circ }$ triangle given as $1:\sqrt{3}:2,$ or $x:x\sqrt{3}:2x.$ Thus, you only need to know the length of one side to find the other two sides in a ${30}^{\circ }\text{-}{60}^{\circ }\text{-}{90}^{\circ }$ triangle.

Example 10.67

Finding Missing Lengths in a $30^{\circ }\text{-}60^{\circ }\text{-}90^{\circ }$ Triangle

Find the measures of the missing lengths of the triangle (Figure 10.145).

Figure 10.145

Solution

We can see that this is a ${30}^{\circ }\text{-}{60}^{\circ }\text{-}{90}^{\circ }$ triangle because we have a right angle and a ${30}^{\circ }$ angle. The remaining angle, therefore, must equal ${60}^{\circ }.$ Because this is a special triangle, we have the ratios of the sides to help us identify the missing lengths. Side $a$ is the shortest side, as it is opposite the smallest angle ${30}^{\circ },$ and we can substitute $a=x.$ The ratios are $x:x\sqrt{3}:2x.$ We have the hypotenuse equaling 10, which corresponds to side $c$, and side $c$ is equal to 2$x$. Now, we must solve for $x$:

$$\begin{array}{ccc}\hfill 2x& \hfill =\hfill & 10\hfill \\ \hfill x& \hfill =\hfill & 5\hfill \end{array}$$

Side $b$ is equal to $x\sqrt{3}$ or $5\sqrt{3}.$ The lengths are $5,5\sqrt{3},10.$

Your Turn 10.67

1.

Find the lengths of the missing sides in the given figure.

Example 10.68

Applying $30^{\circ }\text{-}60^{\circ }\text{-}90^{\circ }$ Triangle to the Real World

A city worker leans a 40-foot ladder up against a building at a ${30}^{\circ }$ angle to the ground (Figure 10.146). How far up the building does the ladder reach?

Figure 10.146

Solution

We have a ${30}^{\circ }\text{-}{60}^{\circ }\text{-}{90}^{\circ }$ triangle, and the hypotenuse is 40 ft. This length is equal to 2$x$, where $x$ is the shortest side. If $2x=40$, then $x=20$. The ladder is leaning on the wall 20 ft up from the ground.

Your Turn 10.68

1.

You want to repair a window on the second floor of your home. If you place the ladder at a ${30}^{\circ <!--\; \circ \; -->}$ angle to the ground, the ladder just about reaches the window. How far from the wall should you place the ladder? How far up will the ladder reach? Make a sketch as an aid.

$45^{\circ }\text{-}45^{\circ }\text{-}90^{\circ }$ Triangles

The ${45}^{\circ }\text{-}{45}^{\circ }\text{-}{90}^{\circ }$ triangle is another special triangle such that with the measure of one side we can find the measures of all the sides. The two angles adjacent to the ${90}^{\circ }$ angle are equal, and each measures ${45}^{\circ }.$ If two angles are equal, so are their opposite sides. The ratio among sides is $1:1:\sqrt{2},$ or $x:x:x\sqrt{2},$ as shown in Figure 10.147.

Figure 10.147 $45^{\circ }\text{-}45^{\circ }\text{-}90^{\circ }$ Triangles

Example 10.69

Finding Missing Lengths of a $45^{\circ }\text{-}45^{\circ }\text{-}90^{\circ }$ Triangle

Find the measures of the unknown sides in the triangle (Figure 10.148).

Figure 10.148

Solution

Because we have a ${45}^{\circ }\text{-}{45}^{\circ }\text{-}{90}^{\circ }$ triangle, we know that the two legs are equal in length and the hypotenuse is a product of one of the legs and $\sqrt{2}.$ One leg measures 3, so the other leg, $a$, measures 3. Remember the ratio of $x:x:x\sqrt{2}.$ Then, the hypotenuse, $c$, equals $3\sqrt{2}.$

Your Turn 10.69

1.

Find the measures of the unknown sides in the given figure.

Trigonometry Functions

Trigonometry developed around 200 BC from a need to determine distances and to calculate the measures of angles in the fields of astronomy and surveying. Trigonometry is about the relationships (or ratios) of angle measurements to side lengths in primarily right triangles. However, trigonometry is useful in calculating missing side lengths and angles in other triangles and many applications.

Trigonometry is based on three functions. We title these functions using the following abbreviations:

• $\sin =\mathrm{sine}$
• $\cos =\mathrm{cosine}$
• $\tan =\mathrm{tangent}$

Letting $r=\sqrt{x^2+y^2},$ which is the hypotenuse of a right triangle, we have Table 10.1. The functions are given in terms of $x$, $y$, and $r$, and in terms of sides relative to the angle, like opposite, adjacent, and the hypotenuse.

We will be applying the sine function, cosine function, and tangent function to find side lengths and angle measurements for triangles we cannot solve using any of the techniques we have studied to this point. In Figure 10.149, we have an illustration mainly to identify $r$ and the sides labeled $x$ and $y$.

Figure 10.149 Angle $\theta$

An angle $\theta$ sweeps out in a counterclockwise direction from the positive $x$-axis and stops when the angle reaches the desired measurement. That ray extending from the origin that marks ${\theta }^{\circ }$ is called the terminal side because that is where the angle terminates. Regardless of the information given in the triangle, we can find all missing sides and angles using the trigonometric functions. For example, in Figure 10.150, we will solve for the missing sides.

Figure 10.150 Solving for Missing Sides

Let’s use the trigonometric functions to find the sides $x$ and $y$. As long as your calculator mode is set to degrees, you do not have to enter the degree symbol. First, let’s solve for $y$.

We have $\sin \theta =\frac{y}{r},$ and $\theta ={60}^{\circ }.$ Then,

Next, let’s find $x$. This is the cosine function. We have $\cos \theta =\frac{x}{r}.$ Then,

Now we have all sides, $1,\sqrt{3},2.$ You can also check the sides using the ${30}^{\circ }\text{-}{60}^{\circ }\text{-}{90}^{\circ }$ ratio of $1:\sqrt{3}:2.$ Table 10.2 is a list of common angles, which you should find helpful.

Example 10.70

Using Trigonometric Functions

Find the lengths of the missing sides for the triangle (Figure 10.151).

Figure 10.151

Solution

We have a ${55}^{\circ }$ angle, and the length of the triangle on the $x$-axis is 6 units.

Step 1: To find the length of $r$, we can use the cosine function, as $\cos \theta =\frac{x}{r}.$ We manipulate this equation a bit to solve for $r$:

$$\begin{array}{ccc}\hfill \cos ({55}^{\circ })& \hfill =\hfill & \frac{6}{r}\hfill \\ \hfill r\cos ({55}^{\circ })& \hfill =\hfill & 6\hfill \\ \hfill r& \hfill =\hfill & \frac{6}{\cos ({55}^{\circ })}\hfill \\ \hfill r& \hfill =\hfill & \frac{6}{0.5736}=10.46\hfill \end{array}$$

Step 2: We can use the Pythagorean Theorem to find the length of $y$. Prove that your answers are correct by using other trigonometric ratios:

$$\begin{array}{ccc}\hfill 6^2+y^2& \hfill =\hfill & {10.46}^2\hfill \\ \hfill y^2& \hfill =\hfill & 109.4-36\hfill \\ \hfill y& \hfill =\hfill & 8.57\hfill \end{array}$$

Step 3: Now that we have $y$, we can use the sine function to prove that $r$ is correct. We have $\sin \theta =\frac{y}{r}.$

$$\begin{array}{ccc}\hfill \sin ({55}^{\circ })& \hfill =\hfill & \frac{8.57}{r}\hfill \\ \hfill r\sin \left({55}^{\circ }\right)& \hfill =\hfill & 8.57\hfill \\ \hfill r& \hfill =\hfill & \frac{8.57}{\sin ({55}^{\circ })}\hfill \\ \hfill & \hfill =\hfill & \frac{8.57}{0.819}=10.46\hfill \end{array}$$

Your Turn 10.70

1.

Find the lengths of the missing sides in the given figure.

To find angle measurements when we have two side measurements, we use the inverse trigonometric functions symbolized as ${\sin }^{-1},$ ${\cos }^{-1},$ or ${\tan }^{-1}.$ The –1 looks like an exponent, but it means inverse. For example, in the previous example, we had $x=6$ and $r=\mathrm{10.46.}$ To find what angle has these values, enter the values for the inverse cosine function ${\cos }^{-1}\left(\frac{x}{r}\right)$ in your calculator:

You can also use the inverse sine function and enter the values of ${\sin }^{-1}\left(\frac{y}{r}\right)$ in your calculator given $y=8.57$ and $r=\mathrm{10.46.}$ We have

Finally, we can also use the inverse tangent function. Recall $\tan \theta =\frac{y}{x}.$ We have

Example 10.71

Solving for Lengths in a Right Triangle

Solve for the lengths of a right triangle in which $\theta ={30}^{\circ }$ and $r=6$ (Figure 10.152).

Figure 10.152

Solution

Step 1: To find side $a$, we use the sine function:

$$\begin{array}{ccc}\hfill \sin {30}^{\circ }& \hfill =\hfill & \frac{a}{6}\hfill \\ \hfill 6\sin {30}^{\circ }& \hfill =\hfill & a=3\hfill \end{array}$$

Step 2: To find $b$, we use the cosine function:

$$\begin{array}{ccc}\hfill \cos {30}^{\circ }& \hfill =\hfill & \frac{b}{6}\hfill \\ \hfill 6\cos {30}^{\circ }& \hfill =\hfill & b=5.196\hfill \end{array}$$

Step 3: Since this is a ${30}^{\circ }\text{-}{60}^{\circ }\text{-}{90}^{\circ }$ triangle and side $b$ should equal $x\sqrt{3},$ if we input 3 for $x$, we have $b=3\sqrt{3}.$ Put this in your calculator and you will get $3\sqrt{3}=\mathrm{5.196.}$

Your Turn 10.71

1.

Find the missing side and angles in the figure shown.

Example 10.72

Finding Altitude

A small plane takes off from an airport at an angle of ${31.3}^{\circ }$ to the ground. About two-thirds of a mile (3,520 ft) from the airport is an 1,100-ft peak in the flight path of the plane (Figure 10.153). If the plane continues that angle of ascent, find its altitude when it is above the peak, and how far it will be above the peak.

Figure 10.153

Solution

To solve this problem, we use the tangent function:

$$\begin{array}{ccc}\hfill \tan {31.3}^{\circ }& \hfill =\hfill & \frac{x}{\mathrm{3,520}}\hfill \\ \hfill \mathrm{3,520}\tan {31.3}^{\circ }& \hfill =\hfill & \mathrm{2,140}\hfill \end{array}$$

The plane’s altitude when passing over the peak is 2,140 ft, and it is 1,040 ft above the peak.

Example 10.73

Finding Unknown Sides and Angles

Suppose you have two known sides, but do not know the measure of any angles except for the right angle (Figure 10.154). Find the measure of the unknown angles and the third side.

Figure 10.154

Solution

Step 1: We can find the third side using the Pythagorean Theorem:

$$\begin{array}{ccc}\hfill 6^2+4^2& \hfill =\hfill & c^2\hfill \\ \hfill 52& \hfill =\hfill & c^2\hfill \\ \hfill 2\sqrt{13}& \hfill =\hfill & c\hfill \end{array}$$

Now, we have all three sides.

Step 2: To find $\theta ,$ we will first find $\sin \theta .$

$$\begin{array}{ccc}\hfill \sin \theta & \hfill =\hfill & \frac{opp}{hyp}\hfill \\ \hfill & \hfill =\hfill & \frac{4}{2\sqrt{13}}\hfill \\ \hfill & \hfill =\hfill & \frac{2}{\sqrt{13}}.\hfill \end{array}$$

The angle $\theta$ is the angle whose sine is $\frac{2}{\sqrt{13}}.$

Step 3: To find $\theta$, we use the inverse sine function:

$$\begin{array}{ccc}\hfill \theta & \hfill =\hfill & {\sin }^{-1}\left(\frac{2}{\sqrt{13}}\right)\hfill \\ \hfill & \hfill =\hfill & {33.7}^{\circ }\hfill \end{array}$$

Step 4: To find the last angle, we just subtract: ${180}^{\circ }-{90}^{\circ }-{33.7}^{\circ }={56.3}^{\circ }$.

Your Turn 10.73

1.

You know the lengths of two sides and the right angle as shown in the figure. Find the length of the third side and the other angles.

Angle of Elevation and Angle of Depression

Other problems that involve trigonometric functions include calculating the angle of elevation and the angle of depression. These are very common applications in everyday life. The angle of elevation is the angle formed by a horizontal line and the line of sight from an observer to some object at a higher level. The angle of depression is the angle formed by a horizontal line and the line of sight from an observer to an object at a lower level.

Example 10.74

Finding the Angle of Elevation

A guy wire of length 110 meters runs from the top of an antenna to the ground (Figure 10.155). If the angle of elevation of an observer to the top of the antenna is ${43}^{\circ },$ how high is the antenna?

Figure 10.155

Solution

We are looking for the height of the tower. This corresponds to the $y$-value, so we will use the sine function:

$$\begin{array}{ccc}\hfill \sin {43}^{\circ }& \hfill =\hfill & \frac{y}{110}\hfill \\ \hfill 110\sin {43}^{\circ }& \hfill =\hfill & y\hfill \\ \hfill 75& \hfill =\hfill & y\hfill \end{array}$$

The tower is 75 m high.

Your Turn 10.74

1.

You travel to Chicago and visit the observation deck at Willis Tower, 1,450 ft above ground. You can see the Magnificent Mile to the northeast 6,864 ft away. What is the angle of depression from the observation deck to the Magnificent Mile?

Example 10.75

Finding Angle of Elevation

You are sitting on the grass flying a kite on a 50-foot string (Figure 10.156). The angle of elevation is ${60}^{\circ }.$ How high above the ground is the kite?

Figure 10.156

Solution

We can solve this using the sine function, $\sin \theta =\frac{opp}{hyp}.$

$$\begin{array}{ccc}\hfill \sin {60}^{\circ }& \hfill =\hfill & \frac{x}{50}\hfill \\ \hfill 50\sin {60}^{\circ }& \hfill =\hfill & x\hfill \\ \hfill & \hfill =\hfill & 43.3\text{ft}\hfill \end{array}$$

Your Turn 10.75

1.

You are flying a kite on a 60-foot string. The angle of elevation from the ground to the kite is ${50}^{\circ <!--\; \circ \; -->}$. How high above the ground is the kite?

Who Knew?

A Visualization of the Pythagorean Theorem

In Figure 10.157, which is one of the more popular visualizations of the Pythagorean Theorem, we see that square $a$ is attached to side $a$; square $b$ is attached to side $b$; and the largest square, square $c$, is attached to side $c$. Side $a$ measures 3 cm in length, side $b$ measures 4 cm in length, and side $c$ measures 5 cm in length. By definition, the area of square $a$ measures 9 square units, the area of square $b$ measures 16 square units, and the area of square $c$ measures 25 square units. Substitute the values given for the areas of the three squares into the Pythagorean Theorem and we have

$$\begin{array}{ccc}\hfill a^2+b^2& \hfill =\hfill & c^2\hfill \\ \hfill 3^2+4^2& \hfill =\hfill & 5^2\hfill \\ \hfill 9+16& \hfill =\hfill & 25\hfill \end{array}$$

Thus, the sum of the squares of the two legs of a right triangle is equal to the square of the hypotenuse, as stated in the Pythagorean Theorem.

Figure 10.157