# 5.10: Law of Sines

**At Grade**Created by: CK-12

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**Practice**Law of Sines

While working in art class you are trying to design pieces of glass that you will eventually fit together into a sculpture. You are drawing out what you think will be a diagram of one of the pieces. You have a side of length 14 inches, and side of length 17 inches, and an angle next to the 17 in side of \begin{align*}35^\circ\end{align*}

It occurs to you that you could use your knowledge of math to find out if you are going to be able to finish the drawing and make a piece that could actually be built.

Can you figure out how to do this?

By the end of this Concept, you'll know how to apply the Law of Sines to determine the number of possible solutions for a triangle.

### Watch This

James Sousa: The Law of Sines: The Ambiguous Case

### Guidance

In \begin{align*}\triangle ABC \end{align*} below, we know two sides and a non-included angle. Remember that the Law of Sines states: \begin{align*}\frac{\sin A}{a} = \frac{\sin B}{b}\end{align*}. Since we know \begin{align*}a, b\end{align*}, and \begin{align*}\angle{A}\end{align*}, we can use the Law of Sines to find \begin{align*}\angle{B}\end{align*}. However, since this is the SSA case, we have to watch out for the Ambiguous case. Since \begin{align*}a < b\end{align*}, we could be faced with situations where either no possible triangles exist, one possible triangle exists, or two possible triangles exist.

To find out how many solutions there are in an ambiguous case, compare the length of \begin{align*}a\end{align*} to \begin{align*}b \sin A\end{align*}. If \begin{align*}a < b \sin A\end{align*}, then there are no solutions. If \begin{align*}a = b \sin A\end{align*}, then there is one solution. If \begin{align*}a > b \sin A\end{align*}, then there are two solutions.

#### Example A

Find \begin{align*}\angle{B}\end{align*}.

**Solution:** Use the Law of Sines to determine the angle.

\begin{align*}\frac{\sin 41}{12} & = \frac{\sin B}{23} \\ 23 \sin 41 & = 12 \sin B \\ \frac{23 \sin 41}{12} & = \sin B \\ 1.257446472 & = \sin B\end{align*}

**Since no angle exists with a sine greater than 1, there is no solution to this problem.**

We also could have compared \begin{align*}a\end{align*} and \begin{align*}b \sin A\end{align*} beforehand to see how many solutions there were to this triangle.

\begin{align*}a = 12, b \sin A = 15.1\end{align*}: since \begin{align*}12 < 15.1, a < b \sin A\end{align*} which tells us there are no solutions.

#### Example B

In \begin{align*}\triangle ABC, a = 15, b = 20\end{align*}, and \begin{align*}\angle{A} = 30^\circ\end{align*}. Find \begin{align*}\angle{B}\end{align*}.

**Solution:** Again in this case, \begin{align*}a < b\end{align*} and we know two sides and a non-included angle. By comparing \begin{align*}a\end{align*} and \begin{align*}b \sin A\end{align*}, we find that \begin{align*}a = 15, b \sin A = 10\end{align*}. Since \begin{align*}15 > 10\end{align*} we know that there will be two solutions to this problem.

\begin{align*}\frac{\sin 30}{15} & = \frac{\sin B}{20} \\ 20 \sin 30 & = 15 \sin B \\ \frac{20 \sin 30}{15} & = \sin B \\ 0.6666667 & = \sin B \\ \angle{B} & = 41.8^\circ\end{align*}

There are two angles less than \begin{align*}180^\circ\end{align*} with a sine of 0.6666667, however. We found the first one, \begin{align*}41.8^\circ\end{align*}, by using the inverse sine function. To find the second one, we will subtract \begin{align*}41.8^\circ\end{align*} from \begin{align*}180^\circ, \angle{B} = 180^\circ - 41.8^\circ = 138.2^\circ\end{align*}.

To check to make sure \begin{align*}138.2^\circ\end{align*} is a solution, we will use the Triangle Sum Theorem to find the third angle. Remember that all three angles must add up to \begin{align*}180^\circ\end{align*}.

\begin{align*}180^\circ - (30^\circ + 41.8^\circ) = 108.2^\circ && \text{or} && 180^\circ - (30^\circ + 138.2^\circ) = 11.8^\circ\end{align*}

This problem yields two solutions. Either \begin{align*}\angle{B} = 41.8^\circ\end{align*} or \begin{align*}138.2^\circ\end{align*}.

#### Example C

A boat leaves lighthouse \begin{align*}A\end{align*} and travels 63km. It is spotted from lighthouse \begin{align*}B\end{align*}, which is 82km away from lighthouse \begin{align*}A\end{align*}. The boat forms an angle of \begin{align*}65.1^\circ\end{align*} with both lighthouses. How far is the boat from lighthouse \begin{align*}B\end{align*}?

**Solution:** In this problem, we again have the SSA angle case. In order to find the distance from the boat to the lighthouse (a) we will first need to find the measure of \begin{align*}\angle A\end{align*}. In order to find \begin{align*}\angle A\end{align*}, we must first use the Law of Sines to find \begin{align*}\angle B\end{align*}. Since \begin{align*}c > b\end{align*}, this situation will yield exactly one answer for the measure of \begin{align*}\angle B\end{align*}.

\begin{align*}\frac{\sin 65.1^\circ}{82} & = \frac{\sin B}{63} \\ \frac{63 \sin 65.1^\circ}{82} & = \sin B \\ 0.6969 & \approx \sin B \\ \angle{B} & = 44.2^\circ\end{align*}

Now that we know the measure of \begin{align*}\angle B\end{align*}, we can find the measure of angle \begin{align*}A, \angle{A} = 180^\circ - 65.1^\circ - 44.2^\circ = 70.7^\circ\end{align*}. Finally, we can use \begin{align*}\angle{A}\end{align*} to find side \begin{align*}a\end{align*}.

\begin{align*}\frac{\sin 65.1^\circ}{82} & = \frac{\sin 70.7^\circ}{a} \\ \frac{82 \sin 70.7^\circ}{\sin 65.1^\circ} & = a \\ a & = 85.3 \end{align*}

The boat is approximately 85.3 km away from lighthouse \begin{align*}B\end{align*}.

### Guided Practice

1. Prove using the Law of Sines: \begin{align*}\frac{a-c}{c} = \frac{\sin A - \sin C}{\sin C}\end{align*}

2. Find all possible measures of angle \begin{align*}B\end{align*} if any exist for the following triangle values: \begin{align*}A = 32.5^\circ, a = 26, b = 37\end{align*}

3. Find all possible measures of angle \begin{align*}B\end{align*} if any exist for the following triangle values: \begin{align*}A = 42.3^\circ, a = 16, b = 26\end{align*}

**Solutions:**

1. \begin{align*}\frac{\sin A}{a} & = \frac{\sin C}{c} \\ c \sin A & = a \sin C \\ c \sin A - c \sin C & = a \sin C - c \sin C \\ c (\sin A - \sin C) & = \sin C (a-c) \\ \frac{\sin A - \sin C}{\sin C} & = \frac{a-c}{c}\end{align*}

2. \begin{align*}\frac{\sin 32.5^\circ}{26} = \frac{\sin B}{37} \rightarrow B = 49.9^\circ\end{align*} or \begin{align*}180^\circ - 49.9^\circ = 130.1^\circ\end{align*}

3. no solution

### Concept Problem Solution

A drawing of this situation looks like this:

You can start by using the Law of Sines:

\begin{align*}\frac{\sin A}{a} = \frac{\sin B}{b}\end{align*}

and substitute known values:

\begin{align*}\frac{\sin 35}{14} = \frac{\sin B}{17}\end{align*}

Then solving for \begin{align*}\sin B\end{align*}:

\begin{align*}\sin B = \frac{17\sin 35^\circ}{14}\end{align*}

And so

\begin{align*}B \approx 44.15^\circ\end{align*}

Since the interior angles of any triangle add up to \begin{align*}180^\circ\end{align*}, we can find \begin{align*}\angle C \end{align*}:

\begin{align*}C = 180^\circ - 35^\circ - 44.15^\circ\\ C = 100.85\\ \end{align*}

This information can be used again in the Law of Sines:

\begin{align*} \frac{\sin A}{a} = \frac{\sin C}{c}\\ \frac{\sin 35}{14} = \frac{\sin 100.86}{c}\\ c = \frac{14 \sin 100.86}{\sin 35}\\ c = \frac{13.75}{.5735}\\ c = 23.976\\ \end{align*}

### Explore More

Find all possible measures of angle \begin{align*}B\end{align*} if any exist for each of the following triangle values.

- \begin{align*}A = 30^\circ, a = 13, b = 15\end{align*}
- \begin{align*}A = 42^\circ, a = 21, b = 12\end{align*}
- \begin{align*}A = 22^\circ, a = 36, b = 37\end{align*}
- \begin{align*}A = 87^\circ, a = 14, b = 12\end{align*}
- \begin{align*}A = 31^\circ, a = 25, b = 44\end{align*}
- \begin{align*}A = 59^\circ, a = 37, b = 41\end{align*}
- \begin{align*}A = 81^\circ, a = 22, b = 20\end{align*}
- \begin{align*}A = 95^\circ, a = 31, b = 34\end{align*}
- \begin{align*}A = 112^\circ, a = 12, b = 15\end{align*}
- \begin{align*}A = 78^\circ, a = 20, b = 16\end{align*}
- In \begin{align*}\triangle ABC\end{align*}, a=10 and \begin{align*}m\angle B=39^\circ\end{align*}. What's a possible value for b that would produce two triangles?
- In \begin{align*}\triangle ABC\end{align*}, a=15 and \begin{align*}m\angle B=67^\circ\end{align*}. What's a possible value for b that would produce no triangles?
- In \begin{align*}\triangle ABC\end{align*}, a=21 and \begin{align*}m\angle B=99^\circ\end{align*}. What's a possible value for b that would produce one triangle?
- Bill and Connie are each leaving for school. Connie's house is 4 miles due east of Bill's house. Bill can see the school in the direction \begin{align*}40^\circ\end{align*} east of north. Connie can see the school on a line \begin{align*}51^\circ\end{align*} west of north. What is the straight line distance of each person from the school?
- Rochelle and Rose are each looking at a hot air balloon. They are standing 2 miles apart. The angle of elevation for Rochelle is \begin{align*}30^\circ\end{align*} and the angle of elevation for Rose is \begin{align*}34^\circ\end{align*}. How high off the ground is the balloon?

ambiguous

Ambiguous means that the given information is not specific. In the context of Geometry or Trigonometry, it means that the given data may not uniquely identify one shape.law of cosines

The law of cosines is a rule relating the sides of a triangle to the cosine of one of its angles. The law of cosines states that , where is the angle across from side .law of sines

The law of sines is a rule applied to triangles stating that the ratio of the sine of an angle to the side opposite that angle is equal to the ratio of the sine of another angle in the triangle to the side opposite that angle.SSA

SSA means side, side, angle and refers to the fact that two sides and the non-included angle of a triangle are known in a problem.### Image Attributions

## Description

## Learning Objectives

Here you'll learn how to apply the Law of Sines when different types of triangles are presented.

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## Date Created:

Sep 26, 2012## Last Modified:

Feb 26, 2015## Vocabulary

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