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Course: Trigonometry > Unit 1

Lesson 4: Solving for an angle in a right triangle using the trigonometric ratios
Math
Trigonometry
Right triangles & trigonometry
Solving for an angle in a right triangle using the trigonometric ratios
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Intro to inverse trig functions

Google Classroom
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?
A right triangle with leg lengths of thirty-five and sixty-five. Angle L is opposite the short leg.and is unknown.
What we know: Relative to ∠L, we know the lengths of the opposite and adjacent sides, so we can write:
tan⁡(L)=oppositeadjacent=3565
But this doesn't help us find the measure of ∠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 ratiosInverse trigonometric functions input side ratios and output angles
sin⁡(θ)=oppositehypotenuse→sin−1⁡(oppositehypotenuse)=θ
cos⁡(θ)=adjacenthypotenuse→cos−1⁡(adjacenthypotenuse)=θ
tan⁡(θ)=oppositeadjacent→tan−1⁡(oppositeadjacent)=θ

Misconception alert!

The expression sin−1⁡(x) is not the same as 1sin⁡(x). In other words, the −1 is not an exponent. Instead, it simply means inverse function.
FunctionGraph
sin⁡(x)
A coordinate plane. The x-axis starts at zero and goes to ninety by tens. It is labeled degrees. The y-axis starts at zero and goes to two by two tenths. It is labeled a ratio. The graphed line is labeled sine of x, which is a nonlinear curve. The line for the sine of x starts at the origin and passes through the points twenty-four, zero point four, forty, zero point sixty-seven, fifty-two, zero point eight, and ninety, one. It is increasing from the origin to the point ninety, one. The rate of change gets smaller, or shallower, as the degrees, or x-values, get larger. All points are approximations.
sin−1⁡(x) (also called arcsin⁡(x)) |
A coordinate plane. The x-axis starts at zero and goes to two by two tenths. It is labeled a ratio. The y-axis starts at zero and goes to ninety by tens. It is labeled degrees. The graphed line is labeled inverse sine of x, which is a nonlinear curve. The line for the inverse sine of x starts at the origin and passes through the points zero point four, twenty-four, zero point sixty-seven, forty, zero point eight, fifty-two, and one, ninety. It is increasing from the origin to the point one, ninety. The rate of change gets larger, or sharper, as the ratios, or x-values, get larger. All points are approximations.
1sin⁡x (also called csc⁡(x)) |
A coordinate plane. The x-axis starts at zero and goes to ninety by tens. It is labeled degrees. The y-axis starts at zero and goes to two by two tenths. It is labeled a ratio. The graphed line is one divided by the sine of x, which is a nonlinear curve. The line for the cosecant of x starts by decreasing from the point thirty, two. It continues decreasing until the point ninety, one. The rate of change starts steep at the point thirty, two, but it get smaller at the graph goes through the points forty, one point fifty-five, fifty, one point three, and sixty-five, one point one. The rate of change is very shallow as the graph approaches the point ninety, one. All points are approximations.
However, there is an alternate notation that avoids this pitfall! We can also express the inverse sine as arcsin, the inverse cosine as arccos, and the inverse tangent as arctan. This notation is common in computer programming languages, and less common 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.
A right triangle with vertices L and V where angle L is unknown. The side between angles L and ninety degrees is sixty-five degress. The side between the right angle and the vertex V is thirty-five units.
m∠L=tan−1⁡( opposite  adjacent)Define.m∠L=tan−1⁡(3565)Substitute values.m∠L≈28.30∘Evaluate with a calculator.

Now let's try some practice problems.

Problem 1
Given △KIP, find m∠I.
Round your answer to the nearest hundredth of a degree.
Right triangle K I P where angle A P I is a right angle. Angle K I P is an unknown angle. K I is ten units. K P is eight units.
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi
∘

Problem 2
Given △DEF, find m∠E.
Round your answer to the nearest hundredth of a degree.
Right triangle D E F where angle D F E is a right angle. Angle D E F is an unknown angle. D F is four units. E F is six units.
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi
∘

Problem 3
Given △LYN, find m∠Y.
Round your answer to the nearest hundredth of a degree.
Right triangle L Y N where angle Y L N is a right angle. Angle L Y N is an unknown angle. Y N is ten units. L Y is three units.
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi
∘

Challenge problem
Solve the triangle completely. That is, find all unknown sides and unknown angles.
Round your answers to the nearest hundredth.
Right Triangle O Z E where angle O E Z is a right angle. Side O Z is nine units. Side E Z is four units.
OE=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi
m∠O=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi
∘
m∠Z=
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3/5
  • a simplified improper fraction, like 7/4
  • a mixed number, like 1 3/4
  • an exact decimal, like 0.75
  • a multiple of pi, like 12 pi or 2/3 pi
∘

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