Topics
5. Triangle Trigonometry
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United States of America
Grades 9-12

5.01 Law of sines

Lesson
Practice

Introduction

We learned how to find missing sides and angles in right triangles in lesson  10.04 Solving right triangles  . In this lesson, we will extend our trigonometry tools to find missing parts of non-right triangles, but we will have to distinguish when this new tool leads to ambiguity in solving problems.

Ratios in oblique triangles

Exploration

Drag the points to create any triangle.

Loading interactive...
  1. Complete the table of values for the triangle created. What do you notice about the relationship between the ratios of the sine of an angle and its opposite side length?
    abcABC\dfrac{\sin A}{a}\dfrac{\sin B}{b}\dfrac{\sin C}{c}
    Triangle values
  2. Create two more triangles (right, obtuse, or acute) and repeat step 1.

The law of sines is a useful equation that relates the sides of a triangle to the sine of the corresponding angles and can be used to solve for missing values in an oblique triangle.

Oblique triangle

Any triangle that is not a right triangle

\displaystyle \dfrac{\sin{A}}{a}=\dfrac{\sin{B}}{b}=\dfrac{\sin{C}}{c}
\bm{A,\,B,\,C}
the measures of the angles in the triangle
\bm{a,\,b,\,c}
the sides in the triangle
Triangle A B C. Side B C, the side opposite angle A, has a length of lowercase a. Side A C, the side opposite angle B, has a length of lowercase b. Side A B, the side opposite angle C, has a length of lowercase c. Arrows from angle A to lowercase a, from angle B to lowercase B, and from angle C to lowercase C, are drawn inside the triangle.

In order to apply the law of sines, we must be given an angle and its opposite side plus one additional side or angle. We will only use two proportions at a time to solve for missing values.

When solving a triangle given two angles and a side, we are guaranteed one unique solution.

Examples

Example 1

Consider the diagram shown below:

a

Formulate a plan for proving the law of sines: \dfrac{\sin A}{a}= \dfrac{\sin B}{b}.

Worked Solution
Create a strategy

The law of sines contains calculations for the sines of angles. In order to calculate the sine of an angle, we need to have a right triangle. So, let's divide \triangle ABC into two right triangles by constructing an altitude \overline{CD} from \angle C to \overline{AB}. We can label the height of the segment as x.

Triangle A B C. Dide C B is labeled as a, side A C is labeled as b, and side A B is labeled as c. A line is drawn perpendicular to the base A B and intersects the base at point D and joins the top vertex C. This segment, C D, is labelled with an x.
Apply the idea

After constructing two right triangles from the given triangle, we can use right triangles and \sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}} to begin proving the law of sines.

b

Use your plan from part (a) to prove the law of sines for any triangle.

Worked Solution
Create a strategy

For \triangle ACD, we can write the trigonometric ratio \sin A = \dfrac{x}{b}. For \triangle BCD, we can write the trigonometric ratio \sin B = \dfrac{x}{a}.

Both equations contain x, so we will start by solving each equation for x and use that to show the law of sines.

Apply the idea

From \triangle ACD:

\displaystyle \sin A\displaystyle =\displaystyle \dfrac{x}{b}Sine of \angle A
\displaystyle b \cdot \sin A\displaystyle =\displaystyle xMultiply both sides of the equation by b

From \triangle BCD:

\displaystyle \sin B\displaystyle =\displaystyle \dfrac{x}{a}Sine of \angle B
\displaystyle a \cdot \sin B\displaystyle =\displaystyle xMultiply both sides of the equation by a

Now, each expression equal to x can be written as follows:

\displaystyle b \cdot \sin A\displaystyle =\displaystyle a \cdot \sin BTransitive property of equality
\displaystyle \dfrac{ \sin A}{a}\displaystyle =\displaystyle \dfrac{ \sin B}{b}Divide both sides of the equation by a and b

Example 2

Find the value of the missing variable for the following triangles. Round to two decimal places.

a
A triangle with two of its angles measuring 104 degrees and 25 degrees. Opposite the 25 degree angle is a side of length 310 centimeters. Opposite the 104 degree angle is a side labeled a.
Worked Solution
Create a strategy

Since we are given an angle and its opposite side plus one additional angle, we can use the law of sines with the given information to write a proportion, then solve for the unknown variable.

Apply the idea
\displaystyle \dfrac{ \sin A}{a}\displaystyle =\displaystyle \dfrac{ \sin B}{b}Law of sines
\displaystyle \dfrac{ \sin 104 \degree}{a}\displaystyle =\displaystyle \dfrac{ \sin 25 \degree}{310}Substitution
\displaystyle 310 \cdot \sin 104 \degree\displaystyle =\displaystyle a \cdot \sin 25 \degreeMultiply both sides of the equation by a and 310
\displaystyle \dfrac{310 \cdot \sin 104 \degree}{\sin 25 \degree}\displaystyle =\displaystyle aDivide both sides of the equation by \sin 25 \degree
\displaystyle 711.73\displaystyle =\displaystyle aEvaluate the multiplication and division

a = 711.73 \text{ cm}

Reflect and check

Since the angles are given in degrees, make sure your calculator is set to degree mode when calculating the trigonometric ratios.

b
A triangle with a side of length 6.58 and opposite angle measuring 116 degrees. Another side has a length of 3.69 and an opposite angle labeled x.
Worked Solution
Apply the idea
\displaystyle \dfrac{ \sin A}{a}\displaystyle =\displaystyle \dfrac{ \sin B}{b}Law of sines
\displaystyle \dfrac{ \sin x}{3.69}\displaystyle =\displaystyle \dfrac{ \sin 116 \degree}{6.58}Substitution
\displaystyle \sin x\displaystyle =\displaystyle \dfrac{3.69 \cdot \sin 116 \degree}{6.58}Multiply both sides of the equation by 3.69
\displaystyle \sin ^{-1} ( \sin x )\displaystyle =\displaystyle \sin ^{-1} \left( \dfrac{3.69 \cdot \sin 116 \degree}{6.58} \right)Apply the inverse trigonometric function to both sides of the equation
\displaystyle x\displaystyle =\displaystyle 30.27 \degreeEvaluate the inverse
Reflect and check

We now have enough information to find the last missing angle of the triangle using the triangle sum theorem:

\displaystyle m \angle A + m \angle B + m \angle C\displaystyle =\displaystyle 180Triangle sum theorem
\displaystyle 116 + 30.27 + m \angle C \displaystyle =\displaystyle 180Substitution
\displaystyle 146.27 + m \angle C\displaystyle =\displaystyle 180Combine like terms
\displaystyle m \angle C\displaystyle =\displaystyle 33.73Subtract 146.27 from both sides of the equation

The final angle in the triangle must be 33.73 \degree.

Example 3

Consider the triangle shown in the figure:

Triangle A B C. Angle B measures 85 degrees and angle C measures 43.2 degrees. Side A C, the side opposite angle B, has a length of 7.8. Side B C, the side opposite angle A, has a length of x.
a

Write the proportion that relates the sides and angles of the triangle using the law of sines.

Worked Solution
Apply the idea

\frac{\sin{A}}{x}=\frac{\sin{85\degree}}{7.8}=\frac{\sin{43.2\degree}}{AB}

Reflect and check

There is no variable label on side \overline{AB} but we can always refer to the measure using the two endpoints.

b

Solve for x.

Worked Solution
Create a strategy

In order to solve for a missing side using the law of sines you need an angle and its opposite side, plus the oppposite angle. Since we know both m\angle{B} and the length of side \overline{AC} we can solve for the value of x.

It may appear at first glance that we don't have enough information because we don't know m\angle{A}, but by applying the triangle sum theorem we can solve for m\angle{A} in order to find x.

Apply the idea

Using the triangle sum theorem: m\angle{A}=180\degree-(85\degree+43.2\degree)=51.8\degree

\displaystyle \frac{a}{\sin{A}}\displaystyle =\displaystyle \frac{b}{\sin{B}}Law of sines
\displaystyle \frac{x}{\sin{51.8\degree}}\displaystyle =\displaystyle \frac{7.8}{\sin{85\degree}}Substitute known values
\displaystyle x\displaystyle =\displaystyle \sin{51.8\degree}\left(\frac{7.8}{\sin{85\degree}}\right)Multiply both sides by \sin{51.8\degree}
\displaystyle x\displaystyle =\displaystyle 6.15Evaluate on a calculator
Reflect and check

The law of sines can be reciprocated, since equivalent ratios can be written in any order, so long as the corresponding portions match. We can see in this example instead of \dfrac{\sin{A}}{a}=\dfrac{\sin{B}}{b}, we used \dfrac{a}{\sin{A}}=\dfrac{b}{\sin{B}}.

Example 4

The Northern lights, or aurora borealis, are a phenomena that occur in the sky where particles in the atmosphere collide and create colorful night skies. Shown are two observation stations that are 28 \text{ mi} apart where scientists observe and photograph the aurora.

Find the height, h, of the aurora borealis and estimate the height to the nearest tenth of a mile.

Worked Solution
Create a strategy

We can use the law of sines and trigonometric ratios in right triangles to estimate the distance from Earth to the Northern lights. Label the diagram with the information given and what we can solve for first:

Apply the idea

By the triangle sum theorem, m \angle B = 180 \degree - \left(50 \degree + 117 \degree \right) = 13 \degree

We can use the law of sines to find missing side x:

\displaystyle \dfrac{\sin A}{a}\displaystyle =\displaystyle \dfrac{\sin B}{b}Law of sines
\displaystyle \dfrac{ \sin 117 \degree}{x}\displaystyle =\displaystyle \dfrac{ \sin 13 \degree}{28}Substitution
\displaystyle 28 \cdot \sin 117 \degree\displaystyle =\displaystyle x \cdot \sin 13 \degreeMultiply both sides of the equation by x and 28
\displaystyle \dfrac{28 \cdot \sin 117 \degree}{\sin 13 \degree}\displaystyle =\displaystyle xDivide both sides of the equation by \sin 13 \degree
\displaystyle 110.9\displaystyle =\displaystyle xEvaluate the multiplication and division

We now know that the distance from Observation Station 1 to the aurora borealis is 110.9 \text{ mi}. If the height from Earth to the aurora borealis is measured as a perpendicular line from Earth, we can construct a right triangle with height, h, and hypotenuse 110.9:

Now, we can use the trigonometric ratio for right triangles \sin \left( \theta \right) = \dfrac{\text{opposite}}{\text{hypotenuse}} to find the missing value of h:

\displaystyle \sin \left( \theta \right)\displaystyle =\displaystyle \dfrac{\text{opposite}}{\text{hypotenuse}}
\displaystyle \sin \left( 50 \degree \right)\displaystyle =\displaystyle \dfrac{h}{110.9}Substitution
\displaystyle 110.9 \cdot \sin \left( 50 \degree \right)\displaystyle =\displaystyle hMultiply both sides of the equation by 110.9
\displaystyle 84.95\displaystyle =\displaystyle hEvaluate the multiplication

The approximate height of the aurora borealis is 85.0 \text{ mi}.

Idea summary

We can apply the law of sines to find missing values in an oblique triangle. Given an angle and its opposite side plus one additional side or angle, we are guaranteed one unique solution:\dfrac{\sin A}{a} = \dfrac{ \sin B}{ b} = \dfrac{ \sin C}{c}

The ambiguous case

Exploration

Consider the triangles shown below.

  1. What do you notice about the triangles?

The ambiguous case occurs when using the law of sines to solve a triangle given two sides and the non-included angle. When solving the ambiguous case it is possible to have no solution, one solution, or two solutions:

  • Case 1: No triangle exists

    This is the case when an error occurs in the calculator while solving for the unknown angle. This means that no such triangle with the given side lengths and angle exists.

  • Case 2: Exactly one triangle exists

    This is the case when the sum of the given angle and the supplement to the solution calculated is equal to or exceeds 180 \degree.

  • Case 3: Two possible triangles exist

    This is the case when the sum of the given angle and the supplement to the solution are less than 180 \degree. Then, the possible triangles are as follows:

    • A triangle that includes the calculated solution
    • A triangle that includes the supplement to the calculated solution

Examples

Example 5

If possible, solve the triangle where a=85, b=93, and m\angle{A}=61\degree.

Worked Solution
Create a strategy

Draw and label a diagram of the triangle; it does not need to be drawn to scale.

Triangle A B C. Angle A measures 61 degrees. Side B C, the side opposite angle A, has a length of 85. Side A C, the side opposite angle B, has a length of 93. A B is drawn as a dashed segment.

From this diagram we can see that we were given two sides and their non-included angle. This is an example of the ambiguous case and when we solve we may find one, two, or no solutions to this problem.

We are given a pair of opposite sides and angles with side a and angle A in addition to side b. We will set up the equation: \frac{\sin{A}}{a}=\frac{\sin{B}}{b} to find m\angle{B}.

Apply the idea
\displaystyle \frac{\sin{A}}{a}\displaystyle =\displaystyle \frac{\sin{B}}{b}Law of sines
\displaystyle \frac{\sin{61 \degree}}{85}\displaystyle =\displaystyle \frac{\sin{B}}{93}Substitute given values
\displaystyle 93\left(\frac{\sin{61 \degree}}{85}\right)\displaystyle =\displaystyle \sin{B}Multiply both sides by 93
\displaystyle \sin^{-1}\left({93\left(\frac{\sin{61 \degree}}{85}\right)}\right)\displaystyle =\displaystyle BApply the inverse sine to both sides
\displaystyle 73.12\degree\displaystyle =\displaystyle BEvaluate

Now that we've found an acute solution to the problem, we need to find its supplement and check if this will produce a valid second solution.

The supplement of \angle B_1=73.12\degree is \angle B_2 =180\degree-73.12\degree=106.88\degree. If we add these two angles together we get 61\degree+106.88\degree=167.88\degree which does not exceed the maximum 180\degree possible in a triangle. That means this triangle has two solutions.

Solution 1: m\angle{B_1}=73.12\degree

So far we have a=85, b=93, m\angle{A}=61\degree and m\angle{B_1}=73.12\degree. Use the triangle angle sum to solve for the missing angle, m\angle{C_1}=180\degree-(61\degree+73.12\degree)=45.88\degree and use the law of sines to find the missing side, c_1:

\displaystyle \frac{a}{\sin{A}}\displaystyle =\displaystyle \frac{c_1}{\sin{C_1}}Law of sines
\displaystyle \frac{85}{\sin{61 \degree}}\displaystyle =\displaystyle \frac{c_1}{\sin{45.88\degree}}Substitute known values
\displaystyle \sin{45.88\degree}\left(\frac{85}{\sin{61 \degree}}\right)\displaystyle =\displaystyle c_1Multiply both sides by \sin{45.88\degree}
\displaystyle 69.77\displaystyle =\displaystyle c_1Evaluate

The three side lengths are a=85, b=93, and c_1=69.77 and the three angles are m\angle{A}=61\degree, m\angle{B_1}=73.12\degree and m\angle{C_1}=45.88\degree.

Solution 2: m\angle{B_2}=106.88\degree

For the second solution we have a=85, b=93, m\angle{A}=61\degree and m\angle{B_2}=106.88\degree. Use the triangle angle sum to solve for the missing angle, m\angle{C_2}=180\degree-(61\degree+106.88\degree)=12.12\degree and use the law of sines to find the missing side, c_2:

\displaystyle \frac{a}{\sin{A}}\displaystyle =\displaystyle \frac{c_2}{\sin{C_2}}Law of sines
\displaystyle \frac{85}{\sin{61 \degree}}\displaystyle =\displaystyle \frac{c_2}{\sin{12.12\degree}}Substitute known values
\displaystyle \sin{12.12\degree}\left(\frac{85}{\sin{61 \degree}}\right)\displaystyle =\displaystyle c_2Multiply both sides by \sin{12.12\degree}
\displaystyle 20.40\displaystyle =\displaystyle c_2Evaluate

The three side lengths are a=85, b=93, and c_2=20.40 and the three angles are m\angle{A}=61\degree, m\angle{B_2}=106.88\degree and m\angle{C_2}=12.12\degree. The diagram shows a visual of both solutions:

Two triangles, triangle A B 1 C 1 and triangle A B 2 C 2. Angle A measures 61 degrees on both triangles. In the first triangle side A C 1 has length of 93, side B 1 C 1 has a length of 85, and side A B 1 has a length of lowercase c 1. In the second triangle side A C 2 has length of 93, side B 2 C 2 has a length of 85, and side A B 2 has a length of lowercase c 2. A B 1 and A B 2 are drawn as dashed segments.
Reflect and check

There will be no solution, or a calculator error, if we attempt to take the inverse sine of a value greater than 1. There will be exactly one solution if the solved angle is acute, but its supplement is too large to create a second valid triangle.

Example 6

Determine how many triangles are possible given a = 29, b = 28, and m \angle A = 37 \degree. Explain your reasoning.

Assume a is the side length opposite \angle A, b is the side length opposite \angle B.

Worked Solution
Create a strategy

Draw and label a diagram of the triangle.

From this diagram we can see that we were given two sides and their non-included angle. This is an example of the ambiguous case and when we solve we may find one, two, or no solutions to this problem.

We are given a pair of opposite sides and angles with side a and angle A in addition to side b. We will set up the equation: \frac{\sin{A}}{a}=\frac{\sin{B}}{b} to find m\angle{B}.

Apply the idea
\displaystyle \frac{\sin{A}}{a}\displaystyle =\displaystyle \frac{\sin{B}}{b}Law of sines
\displaystyle \frac{\sin{37 \degree}}{29}\displaystyle =\displaystyle \frac{\sin{B}}{28}Substitute given values
\displaystyle 28\left(\frac{\sin{37 \degree}}{29}\right)\displaystyle =\displaystyle \sin{B}Multiply both sides by 28
\displaystyle \sin^{-1}\left({28\left(\frac{\sin{37 \degree}}{29}\right)}\right)\displaystyle =\displaystyle BApply the inverse sine to both sides
\displaystyle 35.53\degree\displaystyle =\displaystyle BEvaluate

We will check whether the other possible value for \angle B is valid. The supplement of 35.53 \degree is 180 \degree - 35.53 \degree = 144.47 \degree. Then, the sum of \angle A and 144.47 \degree is 37 \degree + 144.47 \degree = 181.47 \degree. Since this exceeds 180 \degree, there is only one possible triangle from the given information.

Idea summary

The ambiguous case occurs when using the law of sines to solve a triangle given two sides and the non-included angle. Use these rules for determining the possible solution:

  • Case 1: No triangle exists. This is the case when an error occurs in the calculator while solving for the unknown angle. This means that no such triangle with the given side lengths and angle exists
  • Case 2: Exactly one triangle exists. This is the case when the sum of the given angle and the supplement to the solution calculated is equal to or exceeds 180 \degree
  • Case 3: Two possible triangles exist. This is the case when the sum of the given angle and the supplement to the solution are less than 180 \degree. Then, the possible triangles are as follows:
    • A triangle that includes the calculated solution
    • A triangle that includes the supplement to the calculated solution

Outcomes

G.SRT.D.10 (+)

Prove the laws of sines and cosines and use them to solve problems.

G.SRT.D.11 (+)

Understand and apply the law of sines and the law of cosines to find unknown measurements in right and non-right triangles.

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