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, or triangle that does not include a right angle.

\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.

However, 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 where one angle is acute and the other is the obtuse supplement to the acute solution.

An open figure and two triangles. All the three figures have the angle upper case C, the side B C of length lowercase a, and the base drawn as dashed segment. The first figure looks like an incomplete triangle with an open side of length lowercase c that is drawn too short to touch the dashed base that extends to the left of the triangle. The second figure is a right triangle with a vertical leg of length c that touches the dashed base. The third figure is a triangle with a side of length lowercase c that joins vertex B and the dashed base, and it includes a segment drawn from B to another point on the dashed base within the triangle which also has a length of lowercase c.

Worked examples

Example 1

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.

Write the proportions that relate the sides and angles of the triangle using the law of sines.

Solution

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

Reflection

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

Solve for x.

Approach

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.

Solution

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 6.15	Evaluate on a calculator

Reflection

The law of sines can be reciprocated to solve for any missing value. You 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 2

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

Approach

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}.

Solution

\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 B	Apply the inverse sine to both sides
\displaystyle 73.12\degree	\displaystyle =	\displaystyle B	Evaluate

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_1	Multiply both sides by \sin{45.88\degree}
\displaystyle 69.77	\displaystyle =	\displaystyle c_1	Evaluate

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_2	Multiply both sides by \sin{12.12\degree}
\displaystyle 20.40	\displaystyle =	\displaystyle c_2	Evaluate

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.

Reflection

We will get no solution if we attempt to take the inverse sine of a value greater than 1 and we will have exactly one solution if the first solved angle is acute, but its supplement is too large to create a second valid triangle.

Outcomes

MA.912.T.1.2

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

MA.912.T.1.3

Apply the Law of Sines and the Law of Cosines to solve mathematical and real-world problems involving triangles.

Honors: 8.07 Law of sines

Concept summary

Worked examples

Example 1

Solution

Reflection

Approach

Solution

Reflection

Example 2

Approach

Solution

Reflection

Outcomes

MA.912.T.1.2

MA.912.T.1.3

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