<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

Introduction to Law of Sines

Proportion based on ratios of sides and sines of the opposite angles for non-right triangles.

%
Progress
Progress
%
Law of Sines

You know how to use trigonometric ratios to find missing sides in right triangles, but what about non-right triangles? For the triangle below, can you find AB\begin{align*}AB\end{align*}?

Watch This

http://www.youtube.com/watch?v=dxYVBbSXYkA James Sousa: The Law of Sines Basics

Guidance

Look at the triangle below. Based on the angles, can you tell which side is the shortest? Which side is the longest?

The smallest angle is A\begin{align*}\angle A\end{align*}. It opens up to create the shortest side, BC¯¯¯¯¯\begin{align*}\overline{BC}\end{align*}. The largest angle is C\begin{align*}\angle C\end{align*}. It opens up to create the longest side, AB¯¯¯¯¯\begin{align*}\overline{AB}\end{align*}. Clearly, angles and opposite sides within triangles are connected. In fact, the Law of Sines states that the ratio between the sine of an angle and the side opposite the angle is constant for each of the three angle/side pairs within a triangle.

Law of SinessinAa=sinBb=sinCc\begin{align*}\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\end{align*}

In the past, you used the sine ratio to find missing sides and angles in right triangles. With the Law of Sines, you can find missing sides and angles in any type of triangle. In the examples you will learn how to prove the Law of Sines.

Example A

Consider ΔABC\begin{align*}\Delta ABC\end{align*} below. Draw an altitude from vertex B\begin{align*}B\end{align*} that intersects AC¯¯¯¯¯\begin{align*}\overline{AC}\end{align*} to divide ΔABC\begin{align*}\Delta ABC\end{align*} into two right triangles. Find two equations for the length of the altitude.

Solution: Below, the altitude has been drawn and labeled h\begin{align*}h\end{align*}.

Consider the right triangle with hypotenuse of length c\begin{align*}c\end{align*}. h\begin{align*}h\end{align*} is opposite A\begin{align*}\angle A\end{align*} in this triangle. With an opposite side and hypotenuse you can use the sine ratio.

sinAh=hc=csinA

Now consider the right triangle with hypotenuse of length a\begin{align*}a\end{align*}h\begin{align*}h\end{align*} is opposite C\begin{align*}\angle C\end{align*} in this triangle. With an opposite side and hypotenuse you can use the sine ratio.

sinCh=ha=asinC

You have now found two equations for h\begin{align*}h\end{align*}.

Example B

Use your work from Example A to prove the Law of Sines.

Solution: From Example A, h=csinA\begin{align*}h=c \sin A\end{align*} and h=asinC\begin{align*}h=a \sin C\end{align*}. This means that csinA=asinC\begin{align*}c \sin A=a \sin C\end{align*}. Divide both sides by ac\begin{align*}ac\end{align*} and you will see the Law of Sines.

csinAcsinAacsinAa=asinC=asinCac=sinCc

A\begin{align*}A\end{align*} and C\begin{align*}C\end{align*} were two random angles in the original triangle. This proves that in general, within any triangle the ratio of the sine of an angle to its opposite side is constant.

Example C

Use the Law of Sines to find BC\begin{align*}BC\end{align*} and AC\begin{align*}AC\end{align*}.

Solution: Look for an angle and side pair whose measurements are both given. mC=64\begin{align*}m \angle C=64^\circ\end{align*}, and its opposite side is AB=12\begin{align*}AB=12\end{align*}. Next, set up an equation using the two known measurements as one of the ratios in the Law of Sines. Make sure to match angles with opposite sides.

Solve for BC\begin{align*}BC\end{align*}:

sin6412BCBC=sin38BC=12sin38sin648.22

Solve for AC\begin{align*}AC\end{align*}:

sin6412ACAC=sin78AC=12sin78sin6413.06

Concept Problem Revisited

To find AB\begin{align*}AB\end{align*}, you can use the Law of Sines. Match angles with opposite sides.

sin279ABAB=sin72AB=9sin72sin2718.85

Vocabulary

The sine (sin) of an angle within a right triangle is the ratio of the length of the side opposite the angle to the length of the hypotenuse.

The trigonometric ratios are sine, cosine, and tangent.

Trigonometry is the study of triangles.

θ\begin{align*}\theta\end{align*}, or “theta”, is a Greek letter. In geometry, it is often used as a variable to represent an unknown angle measure.

The Law of Sines states that the ratio between the sine of an angle and the side opposite the angle is the same for each of the three angle/side pairs within a triangle.

Law of Sines: sinAa=sinBb=sinCc\begin{align*}\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\end{align*}

Guided Practice

1. The triangle below is drawn to scale. Use the Law of Sines to solve for the measure of angle B\begin{align*}B\end{align*}. Does your answer seem correct?

2. Recall that SSA was not a criterion for triangle congruence. This was because two non-congruent triangles could have the same “side-side-angle” pattern. What does this have to do with the seemingly wrong answer to #2?

3. The triangle below is drawn to scale. Use the Law of Sines to solve for the measure of angle C.

1. From looking at the triangle, B\begin{align*}\angle B\end{align*} looks to be an obtuse angle. Match angles with opposite sides.

sin3614sinBmBmB=sinB22=22sin3614=sin1(22sin3614)67.5

This answer doesn't seem correct because the angle appears to be obtuse (greater than 90\begin{align*}90^\circ\end{align*}).

2. There are two triangles that fit the criteria given in #2 (angle measures have been slightly rounded).

The two possible measures of B\begin{align*}\angle B\end{align*} are supplementary (112+68=180)\begin{align*}(112^\circ+68^\circ=180^\circ)\end{align*}. Also note that sin112=0.927=sin68\begin{align*}\sin 112^\circ=0.927=\sin 68^\circ\end{align*}. The inverse sine function on your calculator will only produce angles between \begin{align*}0^\circ\end{align*} and \begin{align*}90^\circ\end{align*}. In a sense, the calculator was imagining the triangle on the right while you were imagining the triangle on the left. This is a problem when using the Law of Sines to solve for missing angles. If the information you are given fits SSA, it is possible that there are two answers. If your picture is drawn to scale, you can determine whether your answer should be the acute angle or the obtuse angle by looking at the picture.

3. Angle C appears to be obtuse. Match angles with opposite sides.

This is the acute version of the answer. You know the angle should be obtuse, so find the angle supplementary to \begin{align*}65.7^\circ\end{align*}:

\begin{align*}180^\circ-65.7^\circ=114.3^\circ\end{align*}

*Note that \begin{align*}\sin 65.7^\circ=\sin 114.3\end{align*}*

The final answer is: \begin{align*}m \angle C=114.3^\circ\end{align*}.

Practice

For each triangle, find the measure of all missing sides and angles.

1.

2.

3.

4.

5.

6.

7.

8.

9. What does SSA have to do with the Law of Sines? What type of problems do you have to think extra carefully about to make sure you have the correct answer?

10. The triangle below is drawn to scale. Determine the measure of each of the missing angles.

11. The triangle below is drawn to scale. Determine the measure of each of the missing angles.

12. The triangle below is drawn to scale. Determine the measure of each of the missing angles.

13. Use the picture below to derive the Law of Sines.

14. What type of problems can you solve with the Law of Sines?

15. Explain why you cannot use the Law of Sines to solve for \begin{align*}x\end{align*} in the triangle below.

Vocabulary Language: English

law of sines

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.
sine

sine

The sine of an angle in a right triangle is a value found by dividing the length of the side opposite the given angle by the length of the hypotenuse.
Trigonometric Ratios

Trigonometric Ratios

Ratios that help us to understand the relationships between sides and angles of right triangles.