Law of Sines: Formula and Examples

The law of sines says that in any triangle, each side divided by the sine of its opposite angle gives the same number: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ . Use it to find missing sides and angles of triangles that are not right triangles, when you know two angles and a side.

Before this lesson: You should know Solving Right Triangles, Trig Functions of Any Angle and Proportions.

What the Law of Sines Means

In a right triangle, you can use SOH CAH TOA. Most triangles have no right angle, though. These are called oblique triangles. The law of sines works for every triangle, oblique or right.

Label the angles with capital letters $A$ , $B$ and $C$ . Label each side with the lowercase letter of the angle across from it. So side $a$ is across from angle $A$ .

The law says the bigger an angle is, the longer the side across from it, in a fixed ratio. If you know one full pair (a side and the angle across from it), you can find any other side or angle.

Triangle ABC with side a across from angle A, side b across from angle B and side c across from angle C. A dashed orange height h drops from C straight down to side c, making a right angle. — Each side is named for the angle across from it. The height h shows why the law of sines works.

The Law of Sines Formula

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

You can also flip every fraction: $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$ . Put the unknown on top. Use the first form to find a side and the flipped form to find an angle.

Why it works

Drop the height $h$ from $C$ , as in the picture. It splits the triangle into two right triangles. In the left one, $\sin A = \frac{h}{b}$ , so $h = b \sin A$ . In the right one, $\sin B = \frac{h}{a}$ , so $h = a \sin B$ .

Both equal $h$ , so $b \sin A = a \sin B$ . Divide both sides by $\sin A \sin B$ to get $\frac{b}{\sin B} = \frac{a}{\sin A}$ . The same steps with a different height give the $c$ part.

When to use it

AAS: two angles and a side that is not between them.
ASA: two angles and the side between them.
SSA: two sides and an angle across from one of them. This case can give zero, one or two triangles, so it gets its own lesson.

For three sides (SSS) or two sides and the angle between them (SAS), use the law of cosines instead.

Worked Examples

Round sides to the nearest tenth and angles to the nearest tenth of a degree. Use a calculator in degree mode.

Example 1: Find a side (AAS)

In triangle $ABC$ , $A = 40^\circ$ , $B = 65^\circ$ and $a = 12$ . Find $b$ .

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

Write the part of the law with $a$ and $b$ .

\frac{12}{\sin 40^\circ} = \frac{b}{\sin 65^\circ}

Substitute the known values.

b = \frac{12 \sin 65^\circ}{\sin 40^\circ}

Multiply both sides by $\sin 65^\circ$ .

b \approx \frac{12(0.9063)}{0.6428}

Find the sines with a calculator.

b \approx \frac{10.8756}{0.6428}

Multiply on top.

b \approx 16.9

Divide.

Answer: $b \approx 16.9$ . This makes sense: $B$ is bigger than $A$ , so $b$ is longer than $a$ .

Example 2: Solve the whole triangle (ASA)

In triangle $ABC$ , $A = 50^\circ$ , $B = 70^\circ$ and $c = 20$ . Find $C$ , $a$ and $b$ .

C = 180^\circ - 50^\circ - 70^\circ = 60^\circ

Use the triangle angle sum (see triangle angle sum theorem)

\frac{a}{\sin 50^\circ} = \frac{20}{\sin 60^\circ}

Set up a proportion for $a$ .

a = \frac{20 \sin 50^\circ}{\sin 60^\circ} \approx \frac{20(0.7660)}{0.8660}

Multiply both sides by $\sin 50^\circ$ .

a \approx \frac{15.32}{0.8660} \approx 17.7

Simplify.

\frac{b}{\sin 70^\circ} = \frac{20}{\sin 60^\circ}

Set up a proportion for $b$ .

b = \frac{20 \sin 70^\circ}{\sin 60^\circ} \approx \frac{20(0.9397)}{0.8660}

Multiply both sides by $\sin 70^\circ$ .

b \approx \frac{18.794}{0.8660} \approx 21.7

Simplify.

Answer: $C = 60^\circ$ , $a \approx 17.7$ , $b \approx 21.7$ . Use the given side $c = 20$ in each step, not a rounded answer, so errors don’t build up.

Example 3: Find an angle

In triangle $ABC$ , $A = 35^\circ$ , $a = 10$ and $b = 8$ . Find $B$ .

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

Use the flipped form, with the unknown on top.

\frac{\sin B}{8} = \frac{\sin 35^\circ}{10}

Substitute.

\sin B = \frac{8 \sin 35^\circ}{10}

Multiply both sides by 8.

\sin B \approx \frac{8(0.5736)}{10} \approx 0.4589

Find the sine.

B \approx \sin^{-1}(0.4589) \approx 27.3^\circ

Use the inverse sine (see inverse trig functions)

Answer: $B \approx 27.3^\circ$ . Here $a$ is longer than $b$ , so there is only one triangle. When the given angle is across from the shorter side, check for a second triangle, as in the ambiguous case of the law of sines.

Example 4: Word problem with fire towers

Two fire towers, $A$ and $B$ , are 10 miles apart. A ranger at $A$ sees smoke at an angle of $52^\circ$ from the line between the towers. A ranger at $B$ sees it at $61^\circ$ . How far is the fire from tower $A$ ?

F = 180^\circ - 52^\circ - 61^\circ = 67^\circ

Call the fire $F$ . Find the angle at the fire.

The distance from $A$ to the fire is across from the $61^\circ$ angle at $B$ . The 10 miles is across from $F$ .

\frac{d}{\sin 61^\circ} = \frac{10}{\sin 67^\circ}

Set up the proportion.

d = \frac{10 \sin 61^\circ}{\sin 67^\circ}

Multiply both sides by $\sin 61^\circ$ .

d \approx \frac{10(0.8746)}{0.9205}

Find the sines.

d \approx \frac{8.746}{0.9205} \approx 9.5

Simplify.

Answer: The fire is about 9.5 miles from tower $A$ .

Example 5: Height of a tower from two points

From point $P$ , the angle of elevation to the top of a tower is $32^\circ$ . From point $Q$ , 50 meters closer, it is $48^\circ$ . How tall is the tower?

Call the top $T$ . The angle at $Q$ inside triangle $PQT$ is $180^\circ - 48^\circ = 132^\circ$ , because it forms a straight line with the $48^\circ$ angle.

T = 180^\circ - 32^\circ - 132^\circ = 16^\circ

Find the angle at the top.

\frac{QT}{\sin 32^\circ} = \frac{50}{\sin 16^\circ}

Side $QT$ is across from the $32^\circ$ angle at $P$ . Set up the proportion.

QT = \frac{50 \sin 32^\circ}{\sin 16^\circ} \approx 96.13

Multiply both sides by $\sin 32^\circ$ , then use a calculator.

\sin 48^\circ = \frac{h}{QT}

Now use the right triangle at $Q$ (see trigonometric ratios)

h = QT \sin 48^\circ

Multiply both sides by QT.

h \approx 96.13(0.7431) \approx 71.4

Substitute.

Answer: The tower is about 71.4 meters tall. This mixes the law of sines with angles of elevation.

Common Mistakes

Mistake	Example of the error	Fix
Pairing a side with the wrong angle	Using $\frac{a}{\sin B}$	Each side goes with the angle across from it: $a$ with $A$ .
Calculator in radian mode	$\sin 40$ gives about 0.7451	Switch to degree mode: $\sin 40^\circ \approx 0.6428$ .
Using it with SSS or SAS	Trying the law of sines with three sides only	You need one full side-angle pair. Use the law of cosines instead.
Rounding too early	Using a rounded side to find the next side	Use the given values, or keep extra digits until the end.
Forgetting the third angle	Stopping after one side in an ASA problem	Find the third angle first with the angle sum, $180^\circ$ .

Related Methods

Law of cosines

If you know three sides, or two sides and the angle between them, there is no full side-angle pair. The law of cosines handles those cases.

Area with sine

The height $h = b \sin A$ from the proof also gives the area: $\text{Area} = \frac{1}{2} bc \sin A$ . This is useful when you know two sides and the angle between them. See area of oblique triangles.

Right triangles

If $C = 90^\circ$ , then $\sin C = 1$ , so $\frac{a}{\sin A} = c$ . That gives $a = c \sin A$ , the same as sine = opposite over hypotenuse.

Practice Problems

Round sides to the nearest hundredth and angles to the nearest tenth of a degree. Keep the calculator’s full values until the last step.

In triangle $ABC$ , $A = 30^\circ$ , $B = 45^\circ$ and $a = 8$ . Find $b$ .
In triangle $ABC$ , $A = 72^\circ$ , $C = 48^\circ$ and $c = 15$ . Find $a$ .
In triangle $ABC$ , $A = 28^\circ$ , $B = 102^\circ$ and $c = 40$ . Find $C$ , $a$ and $b$ .
In triangle $ABC$ , $A = 30^\circ$ , $B = 45^\circ$ and $a = 6$ . Find the exact value of $b$ .
In triangle $ABC$ , $A = 110^\circ$ , $a = 12$ and $b = 9$ . Find $B$ .
In triangle $ABC$ , $B = 64^\circ$ , $b = 14$ and $c = 10$ . Find $C$ .
In triangle $ABC$ , $B = 36^\circ$ , $C = 80^\circ$ and $a = 25$ . Find $A$ , $b$ and $c$ .
In a right triangle, $C = 90^\circ$ , $c = 10$ and $A = 30^\circ$ . Use the law of sines to find $a$ .
Points $A$ and $B$ are 200 feet apart on one side of a river. A tree $C$ is on the other side. Angle $CAB$ is $54^\circ$ and angle $CBA$ is $71^\circ$ . Find the distance from $A$ to the tree.
From point $P$ , the angle of elevation to the top of a cliff is $25^\circ$ . From point $Q$ , 100 feet closer, it is $40^\circ$ . How tall is the cliff?

Show answers

\frac{b}{\sin 45^\circ} = \frac{8}{\sin 30^\circ}

Set up the proportion.

b = \frac{8 \sin 45^\circ}{\sin 30^\circ}

Multiply both sides by $\sin 45^\circ$ .

b \approx \frac{8(0.7071)}{0.5}

Find the sines.

b \approx \frac{5.6568}{0.5} \approx 11.31

Simplify.

Answer: $b \approx 11.31$

\frac{a}{\sin 72^\circ} = \frac{15}{\sin 48^\circ}

Set up the proportion.

a = \frac{15 \sin 72^\circ}{\sin 48^\circ}

Multiply both sides by $\sin 72^\circ$ .

a \approx \frac{15(0.9511)}{0.7431}

Find the sines.

a \approx \frac{14.2665}{0.7431} \approx 19.20

Simplify.

Answer: $a \approx 19.20$

C = 180^\circ - 28^\circ - 102^\circ = 50^\circ

Use the angle sum to find the third angle.

a = \frac{40 \sin 28^\circ}{\sin 50^\circ} \approx 24.51

Find side $a$ with the law of sines.

b = \frac{40 \sin 102^\circ}{\sin 50^\circ} \approx 51.08

Find side $b$ with the law of sines.

Answer: $C = 50^\circ$ , $a \approx 24.51$ , $b \approx 51.08$

b = \frac{6 \sin 45^\circ}{\sin 30^\circ}

Set up the proportion.

b = \frac{6 \cdot \frac{\sqrt{2}}{2}}{\frac{1}{2}}

Use exact values.

b = \frac{3\sqrt{2}}{\frac{1}{2}}

Simplify the top.

b = 6\sqrt{2}

Dividing by $\frac{1}{2}$ means multiplying by 2.

Answer: $b = 6\sqrt{2} \approx 8.49$

\frac{\sin B}{9} = \frac{\sin 110^\circ}{12}

Use the flipped form.

\sin B = \frac{9 \sin 110^\circ}{12}

Multiply both sides by 9.

\sin B \approx \frac{9(0.9397)}{12} \approx 0.7048

Find the sine.

B \approx \sin^{-1}(0.7048) \approx 44.8^\circ

Take the inverse sine.

Answer: $B \approx 44.8^\circ$ (a triangle can’t have two obtuse angles, so this is the only answer)

\frac{\sin C}{10} = \frac{\sin 64^\circ}{14}

Use the flipped form.

\sin C = \frac{10 \sin 64^\circ}{14}

Multiply both sides by 10.

\sin C \approx \frac{10(0.8988)}{14} \approx 0.6420

Find the sine.

C \approx \sin^{-1}(0.6420) \approx 39.9^\circ

Take the inverse sine.

Answer: $C \approx 39.9^\circ$

A = 180^\circ - 36^\circ - 80^\circ = 64^\circ

Use the angle sum to find the third angle.

b = \frac{25 \sin 36^\circ}{\sin 64^\circ} \approx \frac{25(0.5878)}{0.8988} \approx \frac{14.695}{0.8988} \approx 16.35

Find side $b$ with the law of sines.

c = \frac{25 \sin 80^\circ}{\sin 64^\circ} \approx \frac{25(0.9848)}{0.8988} \approx \frac{24.62}{0.8988} \approx 27.39

Find side $c$ with the law of sines.

Answer: $A = 64^\circ$ , $b \approx 16.35$ , $c \approx 27.39$

\frac{a}{\sin 30^\circ} = \frac{10}{\sin 90^\circ}

Set up the proportion.

\frac{a}{\sin 30^\circ} = 10

Use $\sin 90^\circ = 1$ .

a = 10 \sin 30^\circ

Multiply both sides by $\sin 30^\circ$ .

a = 5

Use $\sin 30^\circ = \frac{1}{2}$ .

Answer: $a = 5$ , which matches sine = opposite over hypotenuse

C = 180^\circ - 54^\circ - 71^\circ = 55^\circ

Angle at the tree.

\frac{AC}{\sin 71^\circ} = \frac{200}{\sin 55^\circ}

$AC$ is across from the $71^\circ$ angle at $B$ .

AC = \frac{200 \sin 71^\circ}{\sin 55^\circ}

Multiply both sides by $\sin 71^\circ$ .

AC \approx 230.85

Use a calculator.

Answer: about 230.85 feet

10.

180^\circ - 40^\circ = 140^\circ

Angle at $Q$ inside the triangle.

T = 180^\circ - 25^\circ - 140^\circ = 15^\circ

Angle at the top.

QT = \frac{100 \sin 25^\circ}{\sin 15^\circ} \approx \frac{100(0.4226)}{0.2588} \approx 163.29

$QT$ is across from the $25^\circ$ angle.

h = QT \sin 40^\circ \approx 163.29(0.6428) \approx 104.96

Right triangle at $Q$ .

Answer: about 104.96 feet

FAQ

What is the law of sines used for?

It finds missing sides and angles in triangles that are not right triangles. You need at least one side and the angle across from it, plus one more side or angle.

When do you use the law of sines vs. the law of cosines?

Use the law of sines for AAS, ASA and SSA, where you know a side-angle pair. Use the law of cosines for SSS and SAS.

Does the law of sines work for right triangles?

Yes. Since $\sin 90^\circ = 1$ , it turns into the usual sine ratio, opposite over hypotenuse.

Why can the law of sines give two answers?

When you solve for an angle, $\sin B$ has the same value for an acute angle and for 180° minus that angle. In the SSA case, both angles can sometimes make a triangle.

Summary

The law of sines: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ .
Each side goes with the angle across from it.
Use it for AAS, ASA and SSA, when you know a full side-angle pair.
Find the third angle with the angle sum, $180^\circ$ .
To find an angle, put the sines on top, then use the inverse sine.

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