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# 1.4: Solving Right Triangles

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Solve right triangles.
• Find the area of any triangle using trigonometry.
• Solve real-world problems that require you to solve a right triangle.
• Find angle measures using inverse trigonometric functions.

## Solving Right Triangles

You can use your knowledge of the Pythagorean Theorem and the six trigonometric functions to solve a right triangle. Because a right triangle is a triangle with a 90 degree angle, solving a right triangle requires that you find the measures of one or both of the other angles. How you solve will depend on how much information is given. The following examples show two situations: a triangle missing one side, and a triangle missing two sides.

Example 1: Solve the triangle shown below.

Solution:

We need to find the lengths of all sides and the measures of all angles. In this triangle, two of the three sides are given. We can find the length of the third side using the Pythagorean Theorem:

\begin{align*}8^2 + b^2 & = 10^2\\ 64 + b^2 & = 100\\ b^2 & = 36\\ b & = \pm 6 \Rightarrow b = 6\end{align*}

(You may have also recognized the “Pythagorean Triple,” 6, 8, 10, instead of carrying out the Pythagorean Theorem.)

You can also find the third side using a trigonometric ratio. Notice that the missing side, \begin{align*}b\end{align*}, is adjacent to \begin{align*}\angle{A}\end{align*}, and the hypotenuse is given. Therefore we can use the cosine function to find the length of \begin{align*}b\end{align*}:

\begin{align*}\cos 53.13^\circ & = \frac{\text{adjacent side}}{\text{hypotenuse}} = \frac{b}{10}\\ 0.6 & = \frac{b}{10}\\ b & = 0.6(10) = 6\end{align*}

We could also use the tangent function, as the opposite side was given. It may seem confusing that you can find the missing side in more than one way. The point is, however, not to create confusion, but to show that you must look at what information is missing, and choose a strategy. Overall, when you need to identify one side of the triangle, you can either use the Pythagorean Theorem, or you can use a trig ratio.

To solve the above triangle, we also have to identify the measures of all three angles. Two angles are given: 90 degrees and 53.13 degrees. We can find the third angle using the Triangle Sum Theorem, \begin{align*}180 - 90 - 53.13 = 36.87^\circ\end{align*}.

Now let’s consider a triangle that has two missing sides.

Example 2: Solve the triangle shown below.

Solution:

In this triangle, we need to find the lengths of two sides. We can find the length of one side using a trig ratio. Then we can find the length of the third side by using a trig ratio with the same given information, not the side we solved for. This is because the side we found is an approximation and would not yield the most accurate answer for the other missing side. Only use the given information when solving right triangles.

We are given the measure of angle \begin{align*}A\end{align*}, and the length of the side adjacent to angle \begin{align*}A\end{align*}. If we want to find the length of the hypotenuse, \begin{align*}c\end{align*}, we can use the cosine ratio:

\begin{align*}\cos 40^\circ & = \frac{adjacent}{hypotenuse} = \frac{6}{c}\\ \cos 40^\circ & = \frac{6}{c}\\ c \cos 40^\circ & = 6\\ c & = \frac{6}{\cos 40^\circ} \approx 7.83\end{align*}

If we want to find the length of the other leg of the triangle, we can use the tangent ratio. This will give us the most accurate answer because we are not using approximations.

\begin{align*}\tan 40^\circ & = \frac{opposite}{adjacent} = \frac{a}{6}\\ a & = 6 \tan 40^\circ \approx 5.03\end{align*}

Now we know the lengths of all three sides of this triangle. In the review questions, you will verify the values of \begin{align*}c\end{align*} and \begin{align*}a\end{align*} using the Pythagorean Theorem. Here, to finish solving the triangle, we only need to find the measure of \begin{align*}\angle{B}: 180 - 90 - 40 = 50^\circ\end{align*}

Notice that in both examples, one of the two non-right angles was given. If neither of the two non-right angles is given, you will need a new strategy to find the angles.

## Inverse Trigonometric Functions

Consider the right triangle below.

From this triangle, we know how to determine all six trigonometric functions for both \begin{align*}\angle{C}\end{align*} and \begin{align*}\angle{T}\end{align*}. From any of these functions we can also find the value of the angle, using our graphing calculators. If you look back at #7 from 1.3, we saw that \begin{align*}\sin 30^\circ = \frac{1}{2}\end{align*}. If you type 30 into your graphing calculator and then hit the \begin{align*}\fbox{\text{SIN}}\end{align*} button, the calculator yields 0.5. (Make sure your calculator’s mode is set to degrees.)

Conversely, with the triangle above, we know the trig ratios, but not the angle. In this case the inverse of the trigonometric function must be used to determine the measure of the angle. These functions are located above the SIN, COS, and TAN buttons on the calculator. To access this function, press \begin{align*}2^{nd}\end{align*} and the appropriate button and the measure of the angle appears on the screen.

\begin{align*}\cos T = \frac{24}{25} \rightarrow \cos^{-1} \frac{24}{25} = T\end{align*} from the calculator we get

Example 3: Find the angle measure for the trig functions below.

a. \begin{align*}\sin x = 0.687\end{align*}

b. \begin{align*}\tan x = \frac{4}{3}\end{align*}

Solution: Plug into calculator.

a. \begin{align*}\sin^{-1} 0.687 = 43.4^\circ\end{align*}

b. \begin{align*}\tan^{-1} \frac{4}{3} = 53.13^\circ\end{align*}

Example 4: You live on a farm and your chore is to move hay from the loft of the barn down to the stalls for the horses. The hay is very heavy and to move it manually down a ladder would take too much time and effort. You decide to devise a make shift conveyor belt made of bed sheets that you will attach to the door of the loft and anchor securely in the ground. If the door of the loft is 25 feet above the ground and you have 30 feet of sheeting, at what angle do you need to anchor the sheets to the ground?

Solution:

From the picture, we need to use the inverse sine function.

\begin{align*}\sin \theta & = \frac{25 \ feet}{30 \ feet}\\ \sin \theta & = 0.8333\\ \sin^{-1} (\sin \theta) & = \sin ^{-1} 0.8333\\ \theta & = 56.4^\circ\end{align*}

The sheets should be anchored at an angle of \begin{align*}56.4^\circ\end{align*}.

## Finding the Area of a Triangle

In Geometry, you learned that the area of a triangle is \begin{align*}A=\frac{1}{2}bh\end{align*}, where \begin{align*}b\end{align*} is the base and \begin{align*}h\end{align*} is the height, or altitude. Now that you know the trig ratios, this formula can be changed around, using sine.

Looking at the triangle above, you can use sine to determine \begin{align*}h, \sin C = \frac{h}{a}\end{align*}. So, solving this equation for \begin{align*}h\end{align*}, we have \begin{align*}a\sin C=h\end{align*}. Substituting this for \begin{align*}h\end{align*}, we now have a new formula for area.

\begin{align*}A = \frac{1}{2} ab \sin C\end{align*}

What this means is you do not need the height to find the area anymore. All you now need is two sides and the angle between the two sides, called the included angle.

Example 5: Find the area of the triangle.

a.

Solution: Using the formula, \begin{align*}A = \frac{1}{2} \ ab \sin C\end{align*}, we have

\begin{align*}A & = \frac{1}{2} \cdot 8 \cdot 13 \cdot \sin 82^\circ\\ & = 4 \cdot 13 \cdot 0.990\\ & = 51.494\end{align*}

Example 6: Find the area of the parallelogram.

Solution: Recall that a parallelogram can be split into two triangles. So the formula for a parallelogram, using the new formula, would be: \begin{align*}A = 2 \cdot \frac{1}{2} \ ab \sin C\end{align*} or \begin{align*}A = ab \sin C\end{align*}.

\begin{align*}A & = 7 \cdot 15 \cdot \sin 65^\circ\\ & = 95.162\end{align*}

## Angles of Elevation and Depression

You can use right triangles to find distances, if you know an angle of elevation or an angle of depression. The figure below shows each of these kinds of angles.

The angle of elevation is the angle between the horizontal line of sight and the line of sight up to an object. For example, if you are standing on the ground looking up at the top of a mountain, you could measure the angle of elevation. The angle of depression is the angle between the horizontal line of sight and the line of sight down to an object. For example, if you were standing on top of a hill or a building, looking down at an object, you could measure the angle of depression. You can measure these angles using a clinometer or a theodolite. People tend to use clinometers or theodolites to measure the height of trees and other tall objects. Here we will solve several problems involving these angles and distances.

Example 7: You are standing 20 feet away from a tree, and you measure the angle of elevation to be \begin{align*}38^\circ\end{align*}. How tall is the tree?

Solution:

The solution depends on your height, as you measure the angle of elevation from your line of sight. Assume that you are 5 feet tall.

The figure shows us that once we find the value of \begin{align*}T\end{align*}, we have to add 5 feet to this value to find the total height of the triangle. To find \begin{align*}T\end{align*}, we should use the tangent value:

\begin{align*}\tan 38^\circ & = \frac{opposite}{adjacent} = \frac{T}{20}\\ \tan 38^\circ & = \frac{T}{20}\\ T & = 20 \tan 38^\circ \approx 15.63\\ \text{Height of tree} & \approx 20.63 \ ft\end{align*}

The next example shows an angle of depression.

Example 8: You are standing on top of a building, looking at a park in the distance. The angle of depression is \begin{align*}53^\circ\end{align*}. If the building you are standing on is 100 feet tall, how far away is the park? Does your height matter?

Solution:

If we ignore the height of the person, we solve the following triangle:

Given the angle of depression is \begin{align*}53^\circ\end{align*}, \begin{align*}\angle{A}\end{align*} in the figure above is \begin{align*}37^\circ\end{align*}. We can use the tangent function to find the distance from the building to the park:

\begin{align*}\tan 37^\circ & = \frac{opposite}{adjacent} = \frac{d}{100}\\ \tan 37^\circ & = \frac{d}{100}\\ d & = 100 \tan 37^\circ \approx 75.36\ ft.\end{align*}

If we take into account the height if the person, this will change the value of the adjacent side. For example, if the person is 5 feet tall, we have a different triangle:

\begin{align*}\tan 37^\circ & = \frac{opposite}{adjacent} = \frac{d}{105}\\ \tan 37^\circ & = \frac{d}{105}\\ d & = 105 \tan 37^\circ \approx 79.12\ ft.\end{align*}

If you are only looking to estimate a distance, then you can ignore the height of the person taking the measurements. However, the height of the person will matter more in situations where the distances or lengths involved are smaller. For example, the height of the person will influence the result more in the tree height problem than in the building problem, as the tree is closer in height to the person than the building is.

## Right Triangles and Bearings

We can also use right triangles to find distances using angles given as bearings. In navigation, a bearing is the direction from one object to another. In air navigation, bearings are given as angles rotated clockwise from the north. The graph below shows an angle of 70 degrees:

It is important to keep in mind that angles in navigation problems are measured this way, and not the same way angles are measured in trigonometry. Further, angles in navigation and surveying may also be given in terms of north, east, south, and west. For example, \begin{align*}N70^\circ E\end{align*} refers to an angle from the north, towards the east, while \begin{align*}N70^\circ W\end{align*} refers to an angle from the north, towards the west. \begin{align*}N70^\circ E\end{align*} is the same as the angle shown in the graph above. \begin{align*}N70^\circ W\end{align*} would result in an angle in the second quadrant.

Example 9: A ship travels on a \begin{align*}N50^\circ E\end{align*} course. The ship travels until it is due north of a port which is 10 nautical miles due east of the port from which the ship originated. How far did the ship travel?

Solution: The angle between \begin{align*}d\end{align*} and 10 nm is the complement of \begin{align*}50^\circ\end{align*}, which is \begin{align*}40^\circ\end{align*}. Therefore we can find \begin{align*}d\end{align*} using the cosine function:

\begin{align*}\cos 40^\circ & = \frac{adjacent}{hypotenuse} = \frac{10}{d}\\ \cos 40^\circ & = \frac{10}{d}\\ d \cos 40^\circ & = 10\\ d & = \frac{10}{\cos 40^\circ} \approx 13.05 \ nm\end{align*}

## Other Applications of Right Triangles

In general, you can use trigonometry to solve any problem that involves right triangles. The next few examples show different situations in which a right triangle can be used to find a length or a distance.

Example 10: The wheelchair ramp

In lesson 3 we introduced the following situation: You are building a ramp so that people in wheelchairs can access a building. If the ramp must have a height of 8 feet, and the angle of the ramp must be about \begin{align*}5^\circ\end{align*}, how long must the ramp be?

Given that we know the angle of the ramp and the length of the side opposite the angle, we can use the sine ratio to find the length of the ramp, which is the hypotenuse of the triangle:

\begin{align*}\sin 5^\circ & = \frac{8}{L}\\ L \sin 5^\circ & = 8\\ L & = \frac{8}{\sin 5^\circ} \approx 91.8 \ ft\end{align*}

This may seem like a long ramp, but in fact a \begin{align*}5^\circ\end{align*} ramp angle is what is required by the Americans with Disabilities Act (ADA). This explains why many ramps are comprised of several sections, or have turns. The additional distance is needed to make up for the small slope.

Right triangle trigonometry is also used for measuring distances that could not actually be measured. The next example shows a calculation of the distance between the moon and the sun. This calculation requires that we know the distance from the earth to the moon. In chapter 5 you will learn the Law of Sines, an equation that is necessary for the calculation of the distance from the earth to the moon. In the following example, we assume this distance, and use a right triangle to find the distance between the moon and the sun.

Example 11: The earth, moon, and sun create a right triangle during the first quarter moon. The distance from the earth to the moon is about 240,002.5 miles. What is the distance between the sun and the moon?

Solution:

Let \begin{align*}d =\end{align*} the distance between the sun and the moon. We can use the tangent function to find the value of \begin{align*}d\end{align*}:

\begin{align*}\tan 89.85^\circ & = \frac{d}{240,002.5}\\ d & = 240,002.5 \tan 89.85^\circ = 91,673,992.71 \ miles\end{align*}

Therefore the distance between the sun and the moon is much larger than the distance between the earth and the moon.

(Source: www.scribd.com, Trigonometry from the Earth to the Stars.)

## Points to Consider

• In what kinds of situations do right triangles naturally arise?
• Are there right triangles that cannot be solved?
• Trigonometry can solve problems at an astronomical scale as well as problems at a molecular or atomic scale. Why is this true?

## Review Questions

1. Solve the triangle.
2. Two friends are writing practice problems to study for a trigonometry test. Sam writes the following problem for his friend Anna to solve: In right triangle \begin{align*}ABC\end{align*}, the measure of angle \begin{align*}C\end{align*} is 90 degrees, and the length of side \begin{align*}c\end{align*} is 8 inches. Solve the triangle. Anna tells Sam that the triangle cannot be solved. Sam says that she is wrong. Who is right? Explain your thinking.
3. Use the Pythagorean Theorem to verify the sides of the triangle in example 2.
4. Estimate the measure of angle \begin{align*}B\end{align*} in the triangle below using the fact that \begin{align*}\sin B = \frac{3}{5}\end{align*} and \begin{align*}\sin 30^\circ = \frac{1}{2}\end{align*}. Use a calculator to find sine values. Estimate \begin{align*}B\end{align*} to the nearest degree.
5. Find the area of the triangle.
6. Find the area of the parallelogram below.
7. The angle of elevation from the ground to the top of a flagpole is measured to be \begin{align*}53^\circ\end{align*}. If the measurement was taken from 15 feet away, how tall is the flagpole?
8. From the top of a hill, the angle of depression to a house is measured to be \begin{align*}14^\circ\end{align*}. If the hill is 30 feet tall, how far away is the house?
9. An airplane departs city A and travels at a bearing of \begin{align*}100^\circ\end{align*}. City B is directly south of city A. When the plane is 200 miles east of city B, how far has the plan traveled? How far apart are City A and City B? What is the length of the slanted outer wall, \begin{align*}w\end{align*}? What is the length of the main floor, \begin{align*}f\end{align*}?
10. A surveyor is measuring the width of a pond. She chooses a landmark on the opposite side of the pond, and measures the angle to this landmark from a point 50 feet away from the original point. How wide is the pond?
11. Find the length of side \begin{align*}x\end{align*}:
12. A deck measuring 10 feet by 16 feet will require laying boards with one board running along the diagonal and the remaining boards running parallel to that board. The boards meeting the side of the house must be cut prior to being nailed down. At what angle should the boards be cut?

1. \begin{align*}\angle{A} & = 50^\circ\\ b & \approx 5.83\\ a & \approx 9.33\end{align*}
2. Anna is correct. There is not enough information to solve the triangle. That is, there are infinitely many right triangles with hypotenuse 8. For example:
3. \begin{align*}6^2 + 5.03^2 = 36 + 25.3009 = 61.3009 = 7.83^2\end{align*}.
4. \begin{align*}\angle{B \approx} 37^\circ\end{align*}
5. \begin{align*}A = \frac{1}{2} \cdot 10 \cdot 12 \cdot \sin 104^\circ = 58.218\end{align*}
6. \begin{align*}A = 4 \cdot 9 \cdot \sin 112^\circ = 33.379\end{align*}
9. The plane has traveled about 203 miles. The two cities are 35 miles apart.
12. \begin{align*}\tan \theta & = \frac{opposite}{adjacent}\\ \tan \theta & = 0.625\\ \theta & = 32^\circ\end{align*}

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