Math Core

So far, every triangle you have solved with trigonometry had a right angle. Most triangles don't. A surveyor measuring across a river, a navigator plotting a course, or a ranger locating a fire usually works with an oblique triangle, one with no right angle. The law of sines is the first of two tools that solve these triangles.

Naming the parts

The standard labels make the formulas easy to read. Name the vertices $A$ , $B$ and $C$ , and use the matching lowercase letter for the side across from each angle: side $a$ is opposite angle $A$ , side $b$ is opposite angle $B$ , and side $c$ is opposite angle $C$ .

In triangle ABC, side a is opposite angle A, side b is opposite angle B, and side c is opposite angle C.

Solving a triangle means finding all three angles and all three sides. You will always be given three of the six parts, at least one of which is a side.

Where the law comes from

Drop an altitude of length $h$ from vertex $C$ to side $c$ . It splits the oblique triangle into two right triangles, and each one gives you an expression for $h$ :

In the right triangle containing angle $A$ , the hypotenuse is $b$ , so $\sin A = \dfrac{h}{b}$ and $h = b \sin A$ .
In the right triangle containing angle $B$ , the hypotenuse is $a$ , so $\sin B = \dfrac{h}{a}$ and $h = a \sin B$ .

Both expressions equal $h$ , so $b \sin A = a \sin B$ . Divide both sides by $\sin A \sin B$ :

\frac{a}{\sin A} = \frac{b}{\sin B}.

Dropping the altitude from a different vertex brings side $c$ into the chain the same way. (The argument still works when one angle is obtuse, because $\sin(180^\circ - \theta) = \sin \theta$ .)

The law of sines

In any triangle $ABC$ ,

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

Equivalently, $\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}$ . Every side is proportional to the sine of the angle across from it.

To use the law, you need one complete pair: a side together with the angle opposite it. Then any other known angle gives its opposite side, and any other known side gives the sine of its opposite angle.

When to use it

The law of sines works directly when you know:

AAS: two angles and a side that is not between them, or
ASA: two angles and the side between them.

In both cases, find the third angle first with $A + B + C = 180^\circ$ . Then you have a complete pair and can find each remaining side with one proportion.

If you know two sides and an angle opposite one of them (SSA), the law of sines still applies, but the answer might be zero, one or two triangles. That is the next lesson. If you know two sides and the angle between them (SAS) or all three sides (SSS), there is no complete pair, so you need the law of cosines instead.

Worked example: AAS: two angles and a side opposite one of them

In triangle $ABC$ , $A = 40^\circ$ , $B = 65^\circ$ and $a = 12$ . Solve the triangle, rounding sides to the nearest tenth.

Two angles and a side not between them (AAS).

Third angle. $C = 180^\circ - 40^\circ - 65^\circ = 75^\circ$ .

Complete pair. You know $a = 12$ and $A = 40^\circ$ , so every ratio equals $\dfrac{12}{\sin 40^\circ}$ .

\begin{aligned} \frac{b}{\sin 65^\circ} &= \frac{12}{\sin 40^\circ} &\quad\Longrightarrow\quad b &= \frac{12 \sin 65^\circ}{\sin 40^\circ} \approx 16.9 \\ \frac{c}{\sin 75^\circ} &= \frac{12}{\sin 40^\circ} &\quad\Longrightarrow\quad c &= \frac{12 \sin 75^\circ}{\sin 40^\circ} \approx 18.0 \end{aligned}

So $C = 75^\circ$ , $b \approx 16.9$ and $c \approx 18.0$ .

Tip

Check the order. The longest side is always across from the largest angle, and the shortest side across from the smallest. Above, $40^\circ < 65^\circ < 75^\circ$ and $12 < 16.9 < 18.0$ , so the answers are consistent.

Worked example: ASA: two angles and the side between them

In triangle $ABC$ , $A = 35^\circ$ , $C = 80^\circ$ and $b = 20$ . Find $a$ and $c$ to the nearest tenth.

Side $b$ lies between angles $A$ and $C$ , and you don't yet know its opposite angle. Find it first: $B = 180^\circ - 35^\circ - 80^\circ = 65^\circ$ . Now $b = 20$ and $B = 65^\circ$ form a complete pair.

a = \frac{20 \sin 35^\circ}{\sin 65^\circ} \approx 12.7, \qquad c = \frac{20 \sin 80^\circ}{\sin 65^\circ} \approx 21.7.

Common mistake

Two mistakes cause most wrong answers here. First, check that your calculator is in degree mode; $\sin 40$ in radian mode is negative. Second, don't round in the middle. Keep the full value of $\dfrac{12}{\sin 40^\circ}$ in your calculator (or type the whole expression at once) and round only the final answer.

Distances you can't measure directly

The law of sines lets you find a distance to something you can't reach, as long as you can measure a baseline and two angles.

Worked example: Locating a fire

Ranger stations $A$ and $B$ are $10$ miles apart along a straight road. From station $A$ , a fire $F$ is seen at an angle of $52^\circ$ from the road; from station $B$ , the angle is $71^\circ$ . How far is the fire from each station, to the nearest tenth of a mile?

Ranger stations A and B are 10 miles apart and both sight the fire F.

The angle at the fire is $F = 180^\circ - 52^\circ - 71^\circ = 57^\circ$ . The road $AB = 10$ is across from $F$ , so $\dfrac{10}{\sin 57^\circ}$ is the complete pair.

Distance from $A$ : side $AF$ is across from angle $B$ , so $AF = \dfrac{10 \sin 71^\circ}{\sin 57^\circ} \approx 11.3$ miles.
Distance from $B$ : side $BF$ is across from angle $A$ , so $BF = \dfrac{10 \sin 52^\circ}{\sin 57^\circ} \approx 9.4$ miles.

The fire is about $11.3$ miles from station $A$ and $9.4$ miles from station $B$ .

Finding an angle

Use the "flipped" form $\dfrac{\sin B}{b} = \dfrac{\sin A}{a}$ when the unknown is an angle. Solve for $\sin B$ , then apply $\sin^{-1}$ . Be careful: $\sin^{-1}$ only returns angles from $0^\circ$ to $90^\circ$ , but a triangle's angle could be obtuse with the same sine. When the known angle is obtuse, the other two angles must be acute, so the calculator's answer is the right one. The general situation is the subject of the next lesson.