The Sine Law

SOH CAH TOA only works in right triangles. The sine law works in any triangle, acute or obtuse. It links each side to the angle across from it, so you can find missing sides and angles in surveying, navigation, and construction problems. All angles are in degrees.

Key ideas

Labelling a triangle

Use capital letters for the angles (vertices) and the matching lower-case letter for the side opposite each one: side $a$ is across from angle $A$, and so on.

The sine law

In any triangle $ABC$:

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

or, flipped (handier when you’re finding an angle):

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

You only ever use two of the three fractions at a time.

When to use it

You need a side and its opposite angle, plus one more piece of information:

• two angles and any side (AAS or ASA): always works, with exactly one triangle
• two sides and an angle opposite one of them (SSA): this can give two triangles; see the ambiguous case

If you know two sides and the angle between them, or all three sides, use the cosine law instead.

The angle sum

The angles in a triangle add to $180^\circ$. If you know two, the third is free.

Worked examples

Example 1: Two angles and the opposite side (AAS)

In $\triangle ABC$, $\angle A = 42^\circ$, $\angle B = 75^\circ$, and $a = 12$ cm. Find $\angle C$, $b$, and $c$ to two decimal places.

Solution. $\angle C = 180^\circ - 42^\circ - 75^\circ = 63^\circ$.

$$\frac{b}{\sin 75^\circ} = \frac{12}{\sin 42^\circ} \quad\Rightarrow\quad b = \frac{12\sin 75^\circ}{\sin 42^\circ} \approx 17.32$$ $$\frac{c}{\sin 63^\circ} = \frac{12}{\sin 42^\circ} \quad\Rightarrow\quad c = \frac{12\sin 63^\circ}{\sin 42^\circ} \approx 15.98$$

So $b \approx 17.32$ cm and $c \approx 15.98$ cm. The longest side, $b$, is opposite the largest angle, $75^\circ$, as it should be.

Example 2: Two angles and the side between them (ASA)

In $\triangle ABC$, $\angle A = 48^\circ$, $\angle B = 70^\circ$, and $c = 25$ m. Find $a$ and $b$.

Solution. First find the angle opposite the known side: $\angle C = 180^\circ - 48^\circ - 70^\circ = 62^\circ$.

$$a = \frac{25\sin 48^\circ}{\sin 62^\circ} \approx 21.04, \qquad b = \frac{25\sin 70^\circ}{\sin 62^\circ} \approx 26.61$$

So $a \approx 21.04$ m and $b \approx 26.61$ m.

Example 3: Finding an angle

In $\triangle ABC$, $a = 15$, $b = 10$, and $\angle A = 70^\circ$. Find $\angle B$ and $\angle C$ to the nearest tenth of a degree.

Solution. Use the flipped form:

$$\frac{\sin B}{10} = \frac{\sin 70^\circ}{15} \quad\Rightarrow\quad \sin B = \frac{10\sin 70^\circ}{15} \approx 0.6265$$

$\angle B \approx \sin^{-1}(0.6265) \approx 38.8^\circ$. (The other angle with this sine, $141.2^\circ$, is impossible: with $70^\circ$ it would add to more than $180^\circ$.)

$\angle C \approx 180^\circ - 70^\circ - 38.8^\circ = 71.2^\circ$.

Example 4: Two lookouts

Two lookouts, $A$ and $B$, are $8$ km apart on a straight shoreline. They spot a boat at angles of $52^\circ$ (at $A$) and $61^\circ$ (at $B$) from the shoreline. How far is the boat from each lookout?

Solution. The angle at the boat $S$ is $180^\circ - 52^\circ - 61^\circ = 67^\circ$, and it’s opposite the known $8$ km side.

$$AS = \frac{8\sin 61^\circ}{\sin 67^\circ} \approx 7.60, \qquad BS = \frac{8\sin 52^\circ}{\sin 67^\circ} \approx 6.85$$

The boat is about $7.60$ km from $A$ and $6.85$ km from $B$.

Common mistakes

Pairing a side with the wrong angle. Each side goes with the angle opposite it. Label the triangle first.

Using the sine law with no matching pair. You need at least one side together with its opposite angle. In Example 2, you had to find $\angle C$ first.

Calculator in radian mode. Check that $\sin 30 = 0.5$ before you start.

Rounding too early. Keep full values in your calculator until the final answer.

Ignoring a second possible angle. When you find an angle with $\sin^{-1}$, there may be an obtuse angle with the same sine. Check whether it could fit.

Practice

1. (Warm-up) Write the sine law for $\triangle PQR$.

Solution

$$\frac{p}{\sin P} = \frac{q}{\sin Q} = \frac{r}{\sin R}$$

2. (Warm-up) In $\triangle ABC$, $\angle A = 30^\circ$, $\angle B = 45^\circ$, and $a = 10$. Find $b$, exactly and to two decimal places.

Solution

$$b = \frac{10\sin 45^\circ}{\sin 30^\circ} = \frac{10\left(\frac{\sqrt{2}}{2}\right)}{\frac{1}{2}} = 10\sqrt{2} \approx 14.14$$

3. (Warm-up) In $\triangle ABC$, $\angle A = 57^\circ$ and $\angle B = 68^\circ$. Find $\angle C$.

Solution

$180^\circ - 57^\circ - 68^\circ = 55^\circ$.

4. (Core) In $\triangle ABC$, $\angle B = 38^\circ$, $\angle C = 100^\circ$, and $b = 15$ m. Find $\angle A$, $a$, and $c$.

Solution

$\angle A = 180^\circ - 38^\circ - 100^\circ = 42^\circ$.

$$a = \frac{15\sin 42^\circ}{\sin 38^\circ} \approx 16.30, \qquad c = \frac{15\sin 100^\circ}{\sin 38^\circ} \approx 23.99$$

So $a \approx 16.30$ m and $c \approx 23.99$ m.

5. (Core) In $\triangle ABC$, $a = 20$, $c = 14$, and $\angle A = 115^\circ$. Find $\angle C$ and $\angle B$.

Solution

$$\sin C = \frac{14\sin 115^\circ}{20} \approx 0.6344 \quad\Rightarrow\quad \angle C \approx 39.4^\circ$$

(The obtuse option, $140.6^\circ$, can’t fit alongside $115^\circ$.)

$\angle B \approx 180^\circ - 115^\circ - 39.4^\circ = 25.6^\circ$.

6. (Core) In $\triangle ABC$, $\angle A = 25^\circ$, $\angle B = 120^\circ$, and $a = 7$. Find $\angle C$, $b$, and $c$.

Solution

$\angle C = 35^\circ$.

$$b = \frac{7\sin 120^\circ}{\sin 25^\circ} \approx 14.34, \qquad c = \frac{7\sin 35^\circ}{\sin 25^\circ} \approx 9.50$$

7. (Core) Points $A$ and $B$ are $120$ m apart on one bank of a river. Point $C$ is on the far bank. $\angle CAB = 72^\circ$ and $\angle CBA = 55^\circ$. How far is it from $A$ to $C$?

Solution

$\angle C = 180^\circ - 72^\circ - 55^\circ = 53^\circ$, opposite $AB$.

$$AC = \frac{120\sin 55^\circ}{\sin 53^\circ} \approx 123.08$$

About $123.08$ m.

8. (Challenge) From point $A$, the angle of elevation to the top $T$ of a building is $32^\circ$. From point $B$, $50$ m closer to the building, it’s $47^\circ$. How tall is the building?

Solution

In $\triangle ABT$: $\angle A = 32^\circ$, $\angle ABT = 180^\circ - 47^\circ = 133^\circ$, so $\angle T = 180^\circ - 32^\circ - 133^\circ = 15^\circ$.

$$BT = \frac{50\sin 32^\circ}{\sin 15^\circ} \approx 102.37$$

Now use the right triangle at $B$: height $= BT\sin 47^\circ \approx 74.87$ m.

9. (Challenge) Prove the sine law for an acute triangle by drawing the altitude $h$ from $C$ to side $c$.

Solution

The altitude splits $\triangle ABC$ into two right triangles. In the left one, $\sin A = \tfrac{h}{b}$, so $h = b\sin A$. In the right one, $\sin B = \tfrac{h}{a}$, so $h = a\sin B$.

Both equal $h$, so $b\sin A = a\sin B$. Dividing by $\sin A\sin B$:

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

Drawing an altitude from a different vertex gives the same result for $c$ and $\angle C$.