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8.05 Solving for angles in right triangles

Lesson

Concept summary

We are already familiar with certain operations having inverse operations that "undo" them - such as addition and subtraction, or multiplication and division. In a similar manner, we can define inverse trigonometric functions to be functions which reverse, or "undo", the existing trigonometric functions.

The functions sine, cosine, and tangent each take an angle measure as input and return a side ratio as output.

For a given angle \theta we have:

\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} \qquad \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} \qquad \tan\theta = \frac{\text{opposite}}{\text{adjacent}}

The inverse functions work in reverse, taking in the ratio of two known sides as input and returning a missing angle measure as output.

So the three inverse trigonometric functions are: \sin^{-1}\left(\frac{\text{opposite}}{\text{hypotenuse}}\right)= \theta \qquad \cos^{-1}\left(\frac{\text{adjacent}}{\text{hypotenuse}}\right) = \theta \qquad \tan^{-1}\left(\frac{\text{opposite}}{\text{adjacent}}\right) = \theta

We can use these inverse trigonometric functions to determine an unknown angle measure when we know a pair of side lengths of a right-triangle.

As these inverse functions "undo" the original functions, we have\sin^{-1}\left(\sin\left(x\right)\right) = x \qquad \cos^{-1}\left(\cos\left(x\right)\right) = x \qquad \tan^{-1}\left(\tan\left(x\right)\right) = xand\sin\left(\sin^{-1}\left(x\right)\right) = x \qquad \cos\left(\cos^{-1}\left(x\right)\right) = x \qquad \tan\left(\tan^{-1}\left(x\right)\right) = x

It is important to note that while the notation used to represent these inverse trigonometric functions looks like a power of -1, these functions are not the same as the reciprocals of the trigonometric functions. That is:

\sin^{-1}(x) \neq \frac{1}{\sin(x)} \qquad \cos^{-1}(x) \neq \frac{1}{\cos(x)} \qquad \tan^{-1}(x) \neq \frac{1}{\tan(x)}

Worked examples

Example 1

If \cos \theta=0.256, find \theta. Round your answer to two decimal places.

Approach

We want to undo the cosine function, that is take the inverse trigonometric functions of both sides to solve for \theta

Solution

\displaystyle \cos \theta\displaystyle =\displaystyle 0.256
\displaystyle \cos^{-1}\left(\cos\left(\theta\right)\right) \displaystyle =\displaystyle \cos^{-1}\left(0.256\right)Apply the inverse function to both sides
\displaystyle \theta\displaystyle =\displaystyle \cos^{-1}\left(0.256\right)Simplify
\displaystyle \theta\displaystyle =\displaystyle 75.17 \degreeEvaluate

\theta=75.17 \degree

Reflection

We need to use a calculator here to evaluate cosine inverse. Make sure your calculator is in degree mode when working with inverse trigonometric ratios.

Example 2

Solve for the value of x. Round your answer to two decimal places.

A right triangle with an angle labelled x degrees. Opposite this angle is side 20 centimeters long. Adjacent to this angle is side 25 centimeters long.

Approach

With respect to x, we are given the opposite side and adjacent side of the right triangle. The opposite side has a length of 20 \text{ cm} and the adjacent side has a length of 25 \text{ cm}. Use this information to set up the appropriate triogonometric ratio and use the correct inverse trigonometric ratio to solve for x.

Solution

\displaystyle \tan x \degree\displaystyle =\displaystyle \dfrac{20}{25}
\displaystyle \tan^{-1}\left(\tan\left(x \degree\right)\right)\displaystyle =\displaystyle \tan^{-1}\left(\dfrac{20}{25}\right)Apply the inverse function to both sides
\displaystyle x \degree\displaystyle =\displaystyle \tan^{-1}\left(\dfrac{20}{25}\right)Simplify
\displaystyle x \degree\displaystyle =\displaystyle 38.66 \degreeEvaluate

x=38.66

Outcomes

MA.912.T.1.1

Define trigonometric ratios for acute angles in right triangles.

MA.912.T.1.2

Solve mathematical and real-world problems involving right triangles using trigonometric ratios and the Pythagorean Theorem.

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