What if you were told that the longest escalator in North America is at the Wheaton Metro Station in Maryland and is 230 feet long and is 115 ft high? What is the angle of elevation, \begin{align*}x^\circ\end{align*}, of this escalator? After completing this Concept, you'll be able use inverse trigonometry to answer this question.
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CK-12 Foundation: Chapter8InverseTrigonometricRatiosA
James Sousa: Introduction to Inverse Trigonometric Functions
Guidance
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.
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 A
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.
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 B
\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*}
Solutions:
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*}
Example C
Solve the right triangle.
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*}
Watch this video for help with the Examples above.
CK-12 Foundation: Chapter8InverseTrigonometricRatiosB
Concept Problem 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*}
Vocabulary
Trigonometry is the study of the relationships between the sides and angles of right triangles. The legs are called adjacent or opposite depending on which acute angle is being used. The three trigonometric (or trig) ratios are sine, cosine, and tangent. The inverse trig ratios, \begin{align*}\sin^{-1}, \ \cos^{-1}\end{align*}, and \begin{align*}\tan^{-1}\end{align*}, allow us to find missing angles when we are given sides.
Guided Practice
1. Solve the right triangle.
2. Solve the right triangle.
Answers:
1. 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*}
2. 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*}
Practice
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.
- \begin{align*}\sin A = 0.5684\end{align*}
- \begin{align*}\cos A =0.1234\end{align*}
- \begin{align*}\tan A = 2.78\end{align*}
Solve the following right triangles. Find all missing sides and angles.
- Writing Explain when to use a trigonometric ratio to find a side length of a right triangle and when to use the Pythagorean Theorem.