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# 8.6: Inverse Trigonometric Ratios

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Use the inverse trigonometric ratios to find an angle in a right triangle.
• Solve a right triangle.
• Apply inverse trigonometric ratios to real-life situation and special right triangles.

## Review Queue

Find the lengths of the missing sides. Round your answer to the nearest hundredth.

1. Draw an isosceles right triangle with legs of length 3. What is the hypotenuse?
2. Use the triangle from #3, to find the sine, cosine, and tangent of \begin{align*}45^\circ\end{align*}.
3. Explain why \begin{align*}\tan 45^\circ = 1\end{align*}.

Know What? The longest escalator in North America is at the Wheaton Metro Station in Maryland. It is 230 feet long and is 115 ft high. What is the angle of elevation, \begin{align*}x^\circ\end{align*}, of this escalator?

## Inverse Trigonometric Ratios

The word inverse is probably familiar to you. In mathematics, once you learn how to do an operation, you also learn how to “undo” it. 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.

When we apply the word inverse to the trigonometric ratios, we can find the acute angle measures within a right triangle. Normally, if you are given an angle and a side of a right triangle, you can find the other two sides, using sine, cosine or tangent. With the inverse trig ratios, you can find the angle measure, given two sides.

Inverse Tangent: If you know the opposite side and adjacent side of an angle in a right triangle, you can use inverse tangent to find the measure of the angle.

Inverse tangent is also called arctangent and is labeled \begin{align*}\tan^{-1}\end{align*} or arctan. The “-1” indicates inverse.

Inverse Sine: If you know the opposite side of an angle and the hypotenuse in a right triangle, you can use inverse sine to find the measure of the angle.

Inverse sine is also called arcsine and is labeled \begin{align*}\sin^{-1}\end{align*} or arcsin.

Inverse Cosine: If you know the adjacent side of an angle and the hypotenuse in a right triangle, you can use inverse cosine to find the measure of the angle.

Inverse cosine is also called arccosine and is labeled \begin{align*}\cos^{-1}\end{align*} or arccos.

Using the triangle below, the inverse trigonometric ratios look like this:

\begin{align*}\tan^{-1} \left ( \frac{b}{a} \right ) & = m \angle B && \tan^{-1} \left ( \frac{a}{b} \right ) = m \angle A\\ \sin^{-1} \left ( \frac{b}{c} \right ) & = m \angle B && \sin^{-1} \left ( \frac{a}{c} \right ) = m \angle A\\ \cos^{-1} \left ( \frac{a}{c} \right ) & = m \angle B && \cos^{-1} \left ( \frac{b}{c} \right ) = m \angle A\end{align*}

In order to actually find the measure of the angles, you will need you use your calculator. On most scientific and graphing calculators, the buttons look like \begin{align*}[\text{SIN}^{-1}], [\text{COS}^{-1}]\end{align*}, and \begin{align*}[\text{TAN}^{-1}]\end{align*}. Typically, you might have to hit a shift or \begin{align*}2^{nd}\end{align*} button to access these functions. For example, on the TI-83 and 84, \begin{align*}[2^{nd}][\text{SIN}]\end{align*} is \begin{align*}[\text{SIN}^{-1}]\end{align*}. Again, make sure the mode is in degrees.

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 tangent inverse tells you the measure of an angle with a given ratio.

Example 1: Use the sides of the triangle and your calculator to find the value of \begin{align*}\angle A\end{align*}. Round your answer to the nearest tenth of a degree.

Solution: In reference to \begin{align*}\angle A\end{align*}, we are given the opposite leg and the adjacent leg. This means we should use the tangent ratio.

\begin{align*}\tan A = \frac{20}{25} = \frac{4}{5}\end{align*}, therefore \begin{align*}\tan^{-1} \left ( \frac{4}{5} \right ) = m \angle A\end{align*}. Use your calculator.

If you are using a TI-83 or 84, the keystrokes would be: \begin{align*}[2^{nd}][\text{TAN}]\left ( \frac{4}{5} \right )\end{align*}[ENTER] and the screen looks like:

So, \begin{align*}m \angle A = 38.7^\circ\end{align*}

Example 2: \begin{align*}\angle A\end{align*} is an acute angle in a right triangle. Use your calculator to find \begin{align*}m \angle A\end{align*} to the nearest tenth of a degree.

a) \begin{align*}\sin A = 0.68\end{align*}

b) \begin{align*}\cos A = 0.85\end{align*}

c) \begin{align*}\tan A = 0.34\end{align*}

Solution:

a) \begin{align*}m \angle A = \sin^{-1} 0.68 = 42.8^\circ\end{align*}

b) \begin{align*}m \angle A = \cos^{-1} 0.85 = 31.8^\circ\end{align*}

c) \begin{align*}m \angle A = \tan^{-1} 0.34 = 18.8^\circ\end{align*}

## Solving Triangles

Now that we know how to use inverse trigonometric ratios to find the measure of the acute angles in a right triangle, we can solve right triangles. To solve a right triangle, you would need to find all sides and angles in a right triangle, using any method. When solving a right triangle, you could use sine, cosine or tangent, inverse sine, inverse cosine, or inverse tangent, or the Pythagorean Theorem. Remember when solving right triangles to only use the values that you are given.

Example 3: Solve the right triangle.

Solution: To solve this right triangle, we need to find \begin{align*}AB, m \angle C\end{align*} and \begin{align*}m \angle B\end{align*}. Use \begin{align*}AC\end{align*} and \begin{align*}CB\end{align*} to give the most accurate answers.

\begin{align*}\underline{AB}\end{align*}: Use the Pythagorean Theorem.

\begin{align*}24^2 + AB^2 & = 30^2\\ 576 + AB^2 & = 900\\ AB^2 & = 324\\ AB & = \sqrt{324} = 18\end{align*}

\begin{align*}\underline{m \angle B}\end{align*}: Use the inverse sine ratio.

\begin{align*}\sin B & = \frac{24}{30} = \frac{4}{5}\\ \sin^{-1} \left ( \frac{4}{5} \right ) & = 53.1^\circ = m \angle B\end{align*}

\begin{align*}\underline{m \angle C}\end{align*}: Use the inverse cosine ratio.

\begin{align*}\cos C & = \frac{24}{30} = \frac{4}{5}\\ \cos^{-1} \left ( \frac{4}{5} \right ) & = 36.9^\circ = m \angle C\end{align*}

Example 4: Solve the right triangle.

Solution: To solve this right triangle, we need to find \begin{align*}AB, BC\end{align*} and \begin{align*}m \angle A\end{align*}.

\begin{align*}\underline{AB}\end{align*}: Use sine ratio.

\begin{align*}\sin 62^\circ & = \frac{25}{AB}\\ AB & = \frac{25}{\sin 62^\circ}\\ AB & \approx 28.31\end{align*}

\begin{align*}\underline{BC}\end{align*}: Use tangent ratio.

\begin{align*}\tan 62^\circ & = \frac{25}{BC}\\ BC & = \frac{25}{\tan 62^\circ}\\ BC & \approx 13.30\end{align*}

\begin{align*}\underline{m \angle A}\end{align*}: Use Triangle Sum Theorem

\begin{align*}62^\circ + 90^\circ + m \angle A & = 180^\circ\\ m \angle A & = 28^\circ\end{align*}

Example 5: Solve the right triangle.

Solution: Even though, there are no angle measures given, we know that the two acute angles are congruent, making them both \begin{align*}45^\circ\end{align*}. Therefore, this is a 45-45-90 triangle. You can use the trigonometric ratios or the special right triangle ratios.

Trigonometric Ratios

\begin{align*}\tan 45^\circ & = \frac{15}{BC} && \sin 45^\circ = \frac{15}{AC}\\ BC & = \frac{15}{\tan 45^\circ} = 15 && \quad \ AC = \frac{15}{\sin 45^\circ} \approx 21.21\end{align*}

45-45-90 Triangle Ratios

\begin{align*}BC = AB = 15, AC = 15 \sqrt{2} \approx 21.21\end{align*}

## Real-Life Situations

Much like the trigonometric ratios, the inverse trig ratios can be used in several real-life situations. Here are a couple examples.

Example 6: A 25 foot tall flagpole casts a 42 feet shadow. What is the angle that the sun hits the flagpole?

Solution: First, draw a picture. The angle that the sun hits the flagpole is the acute angle at the top of the triangle, \begin{align*}x^\circ\end{align*}. From the picture, we can see that we need to use the inverse tangent ratio.

\begin{align*}\tan x & = \frac{42}{25}\\ \tan^{-1} \frac{42}{25} & \approx 59.2^\circ = x\end{align*}

Example 7: Elise is standing on the top of a 50 foot building and spots her friend, Molly across the street. If Molly is 35 feet away from the base of the building, what is the angle of depression from Elise to Molly? Elise’s eye height is 4.5 feet.

Solution: Because of parallel lines, the angle of depression is equal to the angle at Molly, or \begin{align*}x^\circ\end{align*}. We can use the inverse tangent ratio.

\begin{align*}\tan^{-1} \left ( \frac{54.5}{30} \right ) = 61.2^\circ = x\end{align*}

Know What? Revisited To find the escalator’s angle of elevation, we need to use the inverse sine ratio.

\begin{align*}\sin^{-1} \left ( \frac{115}{230} \right ) = 30^\circ \qquad \text{The angle of elevation is}\ 30^\circ.\end{align*}

## Review Questions

Use your calculator to find \begin{align*}m \angle A\end{align*} to the nearest tenth of a degree.

Let \begin{align*}\angle A\end{align*} be an acute angle in a right triangle. Find \begin{align*}m \angle A\end{align*} to the nearest tenth of a degree.

1. \begin{align*}\sin A = 0.5684\end{align*}
2. \begin{align*}\cos A =0.1234\end{align*}
3. \begin{align*}\tan A = 2.78\end{align*}

Solving the following right triangles. Find all missing sides and angles.

1. Writing Explain when to use a trigonometric ratio to find a side length of a right triangle and when to use the Pythagorean Theorem.

Real-Life Situations Use what you know about right triangles to solve for the missing angle. If needed, draw a picture. Round all answers to the nearest tenth of a degree.

1. A 75 foot building casts an 82 foot shadow. What is the angle that the sun hits the building?
2. Over 2 miles (horizontal), a road rises 300 feet (vertical). What is the angle of elevation?
3. A boat is sailing and spots a shipwreck 650 feet below the water. A diver jumps from the boat and swims 935 feet to reach the wreck. What is the angle of depression from the boat to the shipwreck?
4. Elizabeth wants to know the angle at which the sun hits a tree in her backyard at 3 pm. She finds that the length of the tree’s shadow is 24 ft at 3 pm. At the same time of day, her shadow is 6 ft 5 inches. If Elizabeth is 4 ft 8 inches tall, find the height of the tree and hence the angle at which the sunlight hits the tree.
5. Alayna is trying to determine the angle at which to aim her sprinkler nozzle to water the top of a 5 ft bush in her yard. Assuming the water takes a straight path and the sprinkler is on the ground 4 ft from the tree, at what angle of inclination should she set it?
6. Science Connection Would the answer to number 20 be the same every day of the year? What factors would influence this answer? How about the answer to number 21? What factors might influence the path of the water?
7. Tommy was solving the triangle below and made a mistake. What did he do wrong? \begin{align*}\tan^{-1} \left ( \frac{21}{28} \right ) \approx 36.9^\circ\end{align*}
8. Tommy then continued the problem and set up the equation: \begin{align*}\cos 36.9^\circ = \frac{21}{h}\end{align*}. By solving this equation he found that the hypotenuse was 26.3 units. Did he use the correct trigonometric ratio here? Is his answer correct? Why or why not?
9. How could Tommy have found the hypotenuse in the triangle another way and avoided making his mistake?

Examining Patterns Below is a table that shows the sine, cosine, and tangent values for eight different angle measures. Answer the following questions.

\begin{align*}10^\circ\end{align*} \begin{align*}20^\circ\end{align*} \begin{align*}30^\circ\end{align*} \begin{align*}40^\circ\end{align*} \begin{align*}50^\circ\end{align*} \begin{align*}60^\circ\end{align*} \begin{align*}70^\circ\end{align*} \begin{align*}80^\circ\end{align*}
Sine 0.1736 0.3420 0.5 0.6428 0.7660 0.8660 0.9397 0.9848
Cosine 0.9848 0.9397 0.8660 0.7660 0.6428 0.5 0.3420 0.1736
Tangent 0.1763 0.3640 0.5774 0.8391 1.1918 1.7321 2.7475 5.6713
1. What value is equal to \begin{align*}\sin 40^\circ\end{align*}?
2. What value is equal to \begin{align*}\cos 70^\circ\end{align*}?
3. Describe what happens to the sine values as the angle measures increase.
4. Describe what happens to the cosine values as the angle measures increase.
5. What two numbers are the sine and cosine values between?
6. Find \begin{align*}\tan 85^\circ, \tan 89^\circ\end{align*}, and \begin{align*}\tan 89.5^\circ\end{align*} using your calculator. Now, describe what happens to the tangent values as the angle measures increase.
7. Explain why all of the sine and cosine values are less than one. (hint: think about the sides in the triangle and the relationships between their lengths)

1. \begin{align*}\sin 36^\circ = \frac{y}{7} \qquad \cos 36^\circ = \frac{x}{7}\!\\ {\;} \qquad y = 4.11 \qquad \quad \ \ x = 5.66\end{align*}
2. \begin{align*}\cos 12.7^\circ = \frac{40}{x} \qquad \ \tan 12.7^\circ = \frac{y}{40}\!\\ {\;} \qquad \ \ \ x = 41.00 \qquad \qquad \ \ y = 9.01\end{align*}
3. \begin{align*}{\;} \ \sin 45^\circ = \frac{3}{3 \sqrt{2}} = \frac{\sqrt{2}}{2}\!\\ {\;} \ \cos 45^\circ = \frac{3}{3 \sqrt{2}} = \frac{\sqrt{2}}{2}\!\\ {\;} \ \tan 45^\circ = \frac{3}{3} = 1\end{align*}
4. The tangent of \begin{align*}45^\circ\end{align*} equals one because it is the ratio of the opposite side over the adjacent side. In an isosceles right triangle, or 45-45-90 triangle, the opposite and adjacent sides are the same, making the ratio always 1.

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