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Solving for an angle in a right triangle using the trigonometric ratios

Intro to inverse trig functions

Learn about arcsine, arccosine, and arctangent, and how they can be used to solve for a missing angle in right triangles.

Let's take a look at a new type of trigonometry problem. Interestingly, these problems can't be solved with sine, cosine, or tangent.

A problem: In the triangle below, what is the measure of angle

L

What we know: Relative to

\angle L

, we know the lengths of the opposite and adjacent sides, so we can write:

\tan(L) = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{35}{65}

But this doesn't help us find the measure of

\angle L

. We're stuck!

What we need: We need new mathematical tools to solve problems like these. Our old friends sine, cosine, and tangent aren’t up to the task. They take angles and give side ratios, but we need functions that take side ratios and give angles. We need inverse trig functions!

The inverse trigonometric functions

We already know about inverse operations. For example, addition and subtraction are inverse operations, and multiplication and division are inverse operations. Each operation does the opposite of its inverse.

The idea is the same in trigonometry. Inverse trig functions do the opposite of the “regular” trig functions. For example:

Inverse sine $(\sin^{-1})$ does the opposite of the sine.
Inverse cosine $(\cos^{-1})$ does the opposite of the cosine.
Inverse tangent $(\tan^{-1})$ does the opposite of the tangent.

In general, if you know the trig ratio but not the angle, you can use the corresponding inverse trig function to find the angle. This is expressed mathematically in the statements below.

Trigonometric functions input angles and output side ratios		Inverse trigonometric functions input side ratios and output angles
$\sin (\theta)=\dfrac {\text{opposite}}{\text{hypotenuse}}$	$\rightarrow$	$\sin^{-1}\left(\dfrac {\text{opposite}}{\text{hypotenuse}}\right)=\theta$
$\cos (\theta)=\dfrac {\text{adjacent}}{\text{hypotenuse}}$	$\rightarrow$	$\cos^{-1}\left(\dfrac {\text{adjacent}}{\text{hypotenuse}}\right)=\theta$
$\tan (\theta)=\dfrac {\text{opposite}}{\text{adjacent}}$	$\rightarrow$	$\tan^{-1}\left(\dfrac {\text{opposite}}{\text{adjacent}}\right)=\theta$

Misconception alert!

The expression

\sin^{-1}(x)

is not the same as

\dfrac{1}{\sin(x)}

. In other words, the

\small{-1}

is not an exponent. Instead, it simply means inverse function.

I'm confused. Please explain this further.

If a number or variable is raised to the

\small{-1}

power, then this refers to the multiplicative inverse, or the reciprocal. For example,

3^{-1}=\dfrac13

. In general, if

a

is a nonzero real number, then

a^{-1}=\dfrac1a

However, this is not the case for

\sin^{-1}(x)

. This is because the sine is a function, not a quantity!

In general, whenever you see a raised

\small{{-}1}

after a function name, it refers to the inverse function. So, for example, if

f

is a function, then

f^{-1}

represents the inverse of function

f

. The expression

f^{-1}(x)

represents the value of the inverse function for the input

x

However, there is an alternate notation that avoids this pitfall! The inverse sine can also be expressed as

\arcsin

, the inverse cosine as

\arccos

, and the inverse tangent as

\arctan

. This notation is common in computer programming languages, but not in mathematics.

Solving the introductory problem

In the introductory problem, we were given the opposite and adjacent side lengths, so we can use inverse tangent to find the angle.

\begin{aligned} { m\angle L}&=\tan^{-1} \left(\dfrac{\text{} \blueD{\text{ opposite }} }{\text{}\maroonC{\text{ adjacent} }\text{ }} \right)\quad\small{\gray{\text{Define.}}} \\\\ m\angle L&=\tan^{-1}\left(\dfrac{\blueD{35}}{\maroonC{65}}\right)\quad\small{\gray{\text{Substitute values.}}} \\\\ m\angle L &\approx 28.30^\circ \quad\small{\gray{\text{Evaluate with a calculator.}}}\end{aligned}

Now let's try some practice problems.

Problem 1

Given $\triangle KIP$ , find $m\angle I$ .
Round your answer to the nearest hundredth of a degree.

^\circ

Explain

Problem 2

Given $\triangle DEF$ , find $m\angle E$ .
Round your answer to the nearest hundredth of a degree.

^\circ

Explain

Problem 3

Given $\triangle LYN$ , find $m\angle Y$ .
Round your answer to the nearest hundredth of a degree.

^\circ

Explain

Challenge problem

Solve the triangle completely. That is, find all unknown sides and unknown angles.
Round your answers to the nearest hundredth.

OE=

m\angle O =

^\circ

m\angle Z =

^\circ

Explain

Solve for an angle in right triangles

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Solve for an angle in right triangles