What Is the Law of Sines?

Triangle with the law of sines

The unit circle shows us that $\sin^{- 1} (A)$ can be one of two angles in the interval $[0 °, 180 °]$ . So, it’s very important to make sure you’ve gotten the correct angle when using the law of sines.

If the angle you find seems wrong in relation to the figure or the information you’ve been given, try to find $180$ ° minus the angle you found, like this:

180 ° - v

Note! It’s often clever to put the unknown in the top left of these equations. Remember that you only use two of the three terms for any problem! The formula has three terms in it to show that they’re all equal, and that you can use any pair of them.

Formula

The Law of Sines

Given two angles and one side, or two sides and one angle of a triangle $△ A B C$ , then

\begin{align} \frac{a}{\sin A} & = \frac{b}{\sin B} = \frac{c}{\sin C} & (1) \\ \frac{\sin A}{a} & = \frac{\sin B}{b} = \frac{\sin C}{c} & (2) \end{align}

Rule

Uses

You can use the law of sines to

Find a side, if you know two angles and a side that is opposite to one of the known angles—Formula (1).
Find an angle, if you know one angle and two sides, one of which is opposite to the angle you want to find—Formula (2).

Example 1

You have a quadrilateral $□ A B C D$ with $A B = 12$ , $A D = 6$ , $C D = 5$ and $∠ A B D = 30 °$ . Find $∠ A$ and the diagonal $B D$ .

It’s useful to draw an auxiliary figure. It will look like this:

Example of using the law of sines

Begin by finding the angle $∠ A D B$ , to find $∠ A$ : $\begin{array}{l} \frac{\sin ∠ A D B}{12} & = \frac{\sin 30 °}{6} & | \cdot 12 \\ \sin ∠ A D B & = \frac{12 \cdot \sin 30 °}{6} = 1 \\ ∠ A D B & = \sin^{- 1} (1) \\ = 90 ° \\ ∠ A & = 180 ° - 90 ° - 30 ° \\ = 60 ° \end{array}$

Then, to find the diagonal $B D$ , you can either use the Pythagorean Theorem or the law of sines. Let’s look at how to solve it with the law of sines:

\begin{array}{l} \frac{B D}{\sin 60 °} & = \frac{6}{\sin 30 °} & | \cdot \sin 60 ° \\ B D & = \frac{6 \cdot \sin 60 °}{\sin 30 °} \\ \approx \frac{6 \cdot 0.866}{0.5} \\ \approx 10.4 \end{array}

Hence, the diagonal $B D \approx 10.4$ .

Example 2

A triangle $△ A B C$ is defined by $∠ A = 40 °$ , $A C = 8.0 cm$ and $B C = 6.0 cm$ . Draw the triangle, and find the sizes of the remaining sides and angles.

Start by drawing a line segment $l$ and mark a point $A$ . Construct an angle $∠ A = 40 °$ . Then mark $A C = 8.0 cm$ along the left side of $∠ A$ and call that point $C$ . You still don’t know the position of $B$ , but you know that $B C = 6.0 cm$ , so set this as the distance between the legs of your draft compass. Set the point of the draft compass on $C$ and make an arc that intersects $l$ in two points. Call these points $B_{1}$ and $B_{2}$ . This means that there are two triangles that meet the criteria, as you can see in the figure below.