Math Core

Sine, cosine and tangent are defined using right triangles. But most triangles in the real world, like the triangle formed by two lookout towers and a fire, or by three cities on a map, have no right angle. The law of sines extends trigonometry to every triangle.

Notation

In any triangle $ABC$ , the side opposite angle $A$ is called $a$ , the side opposite angle $B$ is $b$ , and the side opposite angle $C$ is $c$ . Lowercase letters are sides, capital letters are angles, and each side is paired with the angle across from it.

Where the law comes from

Draw the altitude $h$ from $C$ to side $c$ . It splits triangle $ABC$ into two right triangles.

The altitude h from C splits triangle ABC into two right triangles.

In the right triangle on the left, $h$ is opposite angle $A$ and $b$ is the hypotenuse, so $\sin A = \dfrac{h}{b}$ , which means $h = b \sin A$ .

In the right triangle on the right, $h$ is opposite angle $B$ and $a$ is the hypotenuse, so $\sin B = \dfrac{h}{a}$ , which means $h = a \sin B$ .

Both expressions equal $h$ , so

$b \sin A = a \sin B \quad\Longrightarrow\quad \frac{a}{\sin A} = \frac{b}{\sin B}.$

Drawing the altitude from a different vertex brings in $c$ and $C$ the same way. (The argument also works for obtuse triangles, where the altitude falls outside, but that proof uses the sine of an obtuse angle, which you'll meet in later courses. Your calculator already knows those values.)

The law of sines

In any triangle $ABC$ ,

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}.$

Each side divided by the sine of its opposite angle gives the same number. You can also flip every fraction: $\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}$ .

When to use it

To use the law of sines you need at least one complete pair: a side together with the angle opposite it. That happens when you know:

AAS (two angles and a side not between them), or
ASA (two angles and the side between them; find the third angle first, and then you have a pair), or
SSA (two sides and an angle opposite one of them).

If you know SAS or SSS, you don't have a complete pair, and you'll need the law of cosines from the next lesson instead.

Worked example: Two angles and a side (AAS)

In triangle $ABC$ , $A = 40^\circ$ , $B = 75^\circ$ and $a = 10$ . Find $b$ and $c$ to the nearest tenth.

First the third angle: $C = 180^\circ - 40^\circ - 75^\circ = 65^\circ$ .

The complete pair is $a = 10$ with $A = 40^\circ$ . Use it every time.

\begin{aligned} \frac{b}{\sin 75^\circ} &= \frac{10}{\sin 40^\circ} &\quad \frac{c}{\sin 65^\circ} &= \frac{10}{\sin 40^\circ} \\ b &= \frac{10 \sin 75^\circ}{\sin 40^\circ} \approx 15.0 &\quad c &= \frac{10 \sin 65^\circ}{\sin 40^\circ} \approx 14.1 \end{aligned}

Check: the largest angle, $B = 75^\circ$ , is opposite the longest side, $b \approx 15.0$ . ✓

Finding an angle

To find an angle, put the sines on top, solve for the sine you want, and then use $\sin^{-1}$ .

Worked example: Two sides and an opposite angle (SSA)

In triangle $ABC$ , $a = 9$ , $b = 7$ and $A = 65^\circ$ . Find $B$ , $C$ and $c$ . Round to the nearest tenth.

\begin{aligned} \frac{\sin B}{7} &= \frac{\sin 65^\circ}{9} \\ \sin B &= \frac{7 \sin 65^\circ}{9} \approx 0.7049 \\ B &\approx \sin^{-1}(0.7049) \approx 44.8^\circ \end{aligned}

Then $C \approx 180^\circ - 65^\circ - 44.8^\circ = 70.2^\circ$ , and

$c = \frac{9 \sin C}{\sin 65^\circ} \approx 9.3.$

(Use the unrounded value of $C$ in your calculator if you can.)

The ambiguous case

SSA is tricky, because the given information doesn't always pin down a single triangle. Picture side $b$ fixed at angle $A$ , and side $a$ hanging from $C$ like a swinging gate. Depending on how long $a$ is, it can reach the base in no place, one place, or two places.

With A = 30°, b = 10 and a = 6, side a can swing to two positions, B₁ and B₂.

The math shows this too. Sines of angles between $0^\circ$ and $180^\circ$ repeat: an angle and its supplement have the same sine, so $\sin \theta = \sin(180^\circ - \theta)$ . For example, $\sin 50^\circ = \sin 130^\circ$ . Your calculator's $\sin^{-1}$ only ever gives the acute one, so you have to check the obtuse one yourself.

Checking SSA

After you find $\sin B$ :

If $\sin B > 1$ : no triangle (side $a$ is too short to reach).
Otherwise let $B_1 = \sin^{-1}(\sin B)$ and $B_2 = 180^\circ - B_1$ .
$B_1$ gives a triangle if $A + B_1 < 180^\circ$ , which is always true when $A$ is acute. $B_2$ gives a second triangle only if $A + B_2 < 180^\circ$ .

If the side opposite the given angle is longer than the other given side ( $a > b$ ), there is exactly one triangle.

Worked example: Two triangles

In triangle $ABC$ , $A = 30^\circ$ , $a = 6$ and $b = 10$ . How many triangles are possible? Find each possible value of $B$ .

$\sin B = \frac{10 \sin 30^\circ}{6} = \frac{10(0.5)}{6} \approx 0.8333$

That's less than $1$ , so there is at least one triangle. $B_1 = \sin^{-1}(0.8333) \approx 56.4^\circ$ .

The supplement is $B_2 = 180^\circ - 56.4^\circ = 123.6^\circ$ . Check: $30^\circ + 123.6^\circ = 153.6^\circ < 180^\circ$ , so there's room for a third angle.

There are two triangles: one with $B \approx 56.4^\circ$ (and $C \approx 93.6^\circ$ ) and one with $B \approx 123.6^\circ$ (and $C \approx 26.4^\circ$ ). These are the triangles $AB_1C$ and $AB_2C$ in the picture.

Common mistake

Always pair each side with the angle opposite it. A common error is to write $\dfrac{a}{\sin B}$ . Before you write a proportion, check that each fraction has matching letters: $a$ with $A$ , $b$ with $B$ , $c$ with $C$ .

Practice

Practice 1

In triangle $ABC$ , $A = 30^\circ$ , $B = 45^\circ$ and $a = 8$ . Find the exact value of $b$ . (Type a square root as sqrt(…).)