The Law of Sines, and the Ambiguous Case

Q: What is the Law of Sines?

In any triangle, a/sin A = b/sin B = c/sin C, where each lowercase side is opposite the uppercase angle with the same letter.

Q: When do I use the Law of Sines?

When a side and the angle opposite it are both known, plus one more side or angle. Two angles and any side always work.

Q: What is the ambiguous case?

Two sides and an angle that is not between them. The angle opposite the other side has two possible values with the same sine, so there may be two triangles, one, or none.

Q: How do I prove the Law of Sines?

Drop the height from one vertex. It equals b sin A using one right triangle and a sin B using the other, so b sin A = a sin B, which rearranges to a/sin A = b/sin B.

Q: Does it work for right triangles?

Yes. With C = 90°, sin C = 1, and a/sin A = c reduces to sin A = a/c, the ordinary sine ratio.

Quick answer

In any triangle, not only a right one, each side divided by the sine of the angle opposite it gives the same number. Dropping one height from a vertex proves it, since that height can be written two ways. The law finds missing sides when two angles and a side are known. Given two sides and an angle that is not between them, it can return two different triangles, or none.

What you'll learn

Prove the Law of Sines by drawing an altitude
Solve a triangle given two angles and a side
Decide whether two sides and a non-included angle give zero, one or two triangles

Triangles without a right angle

Sine, cosine and tangent solve right triangles. Most triangles have no right angle, and SOH CAH TOA has nothing to say about them directly.

Label any triangle the standard way: angles $A$ , $B$ , $C$ at the vertices, and each side named by the lowercase letter of the angle opposite it. Then:

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

That is the Law of Sines. Each side, divided by the sine of the angle across from it, gives the same number.

Why the ratios match

Drop the height $h$ from $C$ to side $AB$ , meeting it at $D$ . The height splits the triangle into two right triangles, and each one gives $h$ a different name.

One height, written two ways

In right triangle $ADC$ , the hypotenuse is $b$ and $h$ is opposite angle $A$ :

h = b\sin A

In right triangle $BDC$ , the hypotenuse is $a$ and $h$ is opposite angle $B$ :

h = a\sin B

It is the same height, so the two expressions are equal. Divide both sides by $\sin A \sin B$ :

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

Dropping the height from $A$ instead gives $\tfrac{b}{\sin B} = \tfrac{c}{\sin C}$ the same way, so all three ratios agree. If an angle is obtuse the height lands outside the triangle, and the argument still holds because $\sin(180° - B) = \sin B$ .

The area formula gives a second proof. The area can be written with any pair of sides:

\tfrac{1}{2}bc\sin A = \tfrac{1}{2}ac\sin B = \tfrac{1}{2}ab\sin C

Dividing all three by $\tfrac{1}{2}abc$ leaves $\tfrac{\sin A}{a} = \tfrac{\sin B}{b} = \tfrac{\sin C}{c}$ , which is the same law turned upside down.

Two angles and a side

Given $A = 40°$ , $B = 65°$ and $a = 10$ , solve the triangle.

The third angle comes from the angle sum:

C = 180° - 40° - 65° = 75°

Each unknown side sits in a ratio with the known pair $\tfrac{a}{\sin A}$ :

b = \frac{10\sin 65°}{\sin 40°} \approx 14.10 \qquad c = \frac{10\sin 75°}{\sin 40°} \approx 15.03

These are the measurements of the triangle in the figure above.

The ambiguous case

Now give two sides and an angle that is not between them: $A = 30°$ , $a = 6$ and $b = 10$ . Find $B$ .

\sin B = \frac{b\sin A}{a} = \frac{10 \cdot 0.5}{6} \approx 0.8333

A calculator returns $B \approx 56.44°$ . But $123.56°$ has the same sine, since $\sin(180° - B) = \sin B$ . Both fit: $30° + 123.56°$ is still under $180°$ .

There are two triangles. The picture shows why. Side $a = 6$ hangs from $C$ and can swing to meet the base line in two places.

Two triangles from the same SSA information

$\sin B$ from the Law of Sines	Triangles
greater than $1$	none — side $a$ is too short to reach
exactly $1$	one, with a right angle at $B$
less than $1$ , and the obtuse $B$ still fits	two
less than $1$ , but the obtuse $B$ does not fit	one

The obtuse option “fits” when $A + (180° - B) < 180°$ , which is the same as $A < B$ for the acute value of $B$ .

Worked examples

Common mistakes

Practice problems

Given $A = 50°$ , $B = 60°$ and $a = 12$ , find $C$ , $b$ and $c$ .

Answer

$C = 70°$ , $b \approx 13.57$ , $c \approx 14.72$

Full solution

$C = 180° - 50° - 60° = 70°$ .

$b = \tfrac{12\sin 60°}{\sin 50°} \approx 13.57$ and $c = \tfrac{12\sin 70°}{\sin 50°} \approx 14.72$ .
Given $A = 35°$ , $C = 85°$ and $c = 20$ , find $b$ .

Answer

$b \approx 17.39$

Full solution

$B = 180° - 35° - 85° = 60°$ , so $b = \tfrac{20\sin 60°}{\sin 85°} \approx 17.39$ .
How many triangles have $A = 30°$ , $a = 5$ and $b = 10$ ?

Answer

One, a right triangle.

Full solution

$\sin B = \tfrac{10 \cdot 0.5}{5} = 1$ , so $B = 90°$ . Side $a$ reaches the base line at exactly one point, the foot of the height from $C$ .
How many triangles have $A = 30°$ , $a = 4$ and $b = 10$ ?

Answer

None.

Full solution

$\sin B = 1.25$ , which no angle has.
Given $A = 30°$ , $a = 6$ and $b = 10$ , find both possible values of $B$ and of $C$ .

Answer

$B \approx 56.44°$ with $C \approx 93.56°$ , or $B \approx 123.56°$ with $C \approx 26.44°$ .

Full solution

$\sin B \approx 0.8333$ gives $B \approx 56.44°$ or $180° - 56.44° = 123.56°$ .

Each leaves room for a positive third angle: $180° - 30° - 56.44° = 93.56°$ and $180° - 30° - 123.56° = 26.44°$ .
Given $A = 35°$ , $a = 9$ and $b = 12$ , how many triangles are there?

Answer

Two.

Full solution

$\sin B = \tfrac{12\sin 35°}{9} \approx 0.7648$ , so $B \approx 49.89°$ or $130.11°$ .

With the obtuse value, $35° + 130.11° = 165.11°$ , still under $180°$ . Both triangles exist.
Two points on a riverbank are $150$ feet apart. A post across the river makes angles of $75°$ and $55°$ with the bank at the two points. How far is the post from the point with the $75°$ angle?

Answer

About $160.4$ feet

Full solution

The angle at the post is $180° - 75° - 55° = 50°$ .

The distance from the $75°$ point is the side opposite the $55°$ angle: $\tfrac{150\sin 55°}{\sin 50°} \approx 160.4$ feet.
Use the area formula to show that $\tfrac{a}{\sin A} = \tfrac{b}{\sin B}$ .

Answer

Set $\tfrac{1}{2}bc\sin A = \tfrac{1}{2}ac\sin B$ and simplify.

Full solution

Both expressions give the area of the same triangle, so they are equal.

Cancel $\tfrac{1}{2}c$ from both sides: $b\sin A = a\sin B$ . Divide by $\sin A \sin B$ : $\tfrac{b}{\sin B} = \tfrac{a}{\sin A}$ .
Show that the Law of Sines agrees with the sine ratio when $C = 90°$ .

Answer

With $\sin C = 1$ , the law says $\tfrac{a}{\sin A} = c$ , so $\sin A = \tfrac{a}{c}$ .

Full solution

In a right triangle with the right angle at $C$ , side $c$ is the hypotenuse.

$\tfrac{a}{\sin A} = \tfrac{c}{\sin 90°} = c$ rearranges to $\sin A = \tfrac{a}{c}$ : opposite over hypotenuse, the definition from right-triangle trigonometry.
Given $A = 30°$ , $a = 6$ and $b = 10$ , Nora finds $B \approx 56.4°$ and solves the triangle. Find what she missed.

Hint

Which other angle has the same sine?

Answer

A second triangle, with $B \approx 123.6°$ .

Full solution

The inverse sine gives only the acute angle. $180° - 56.4° = 123.6°$ has the same sine, and $30° + 123.6° = 153.6°$ leaves room for a third angle.

So two different triangles match the given information, and the answer needs both. Swinging the side of length $6$ from $C$ meets the base line in two places, one for each triangle.

Frequently asked questions

What is the Law of Sines?

In any triangle, a/sin A = b/sin B = c/sin C, where each lowercase side is opposite the uppercase angle with the same letter.

When do I use the Law of Sines?

When a side and the angle opposite it are both known, plus one more side or angle. Two angles and any side always work.

What is the ambiguous case?

Two sides and an angle that is not between them. The angle opposite the other side has two possible values with the same sine, so there may be two triangles, one, or none.

How do I prove the Law of Sines?

Drop the height from one vertex. It equals b sin A using one right triangle and a sin B using the other, so b sin A = a sin B, which rearranges to a/sin A = b/sin B.

Does it work for right triangles?

Yes. With C = 90°, sin C = 1, and a/sin A = c reduces to sin A = a/c, the ordinary sine ratio.

What to learn next

The Law of CosinesWhy c² = a² + b² − 2ab cos C is the Pythagorean theorem with a correction, proved with coordinates, and how to find a side from SAS or an angle from SSS.

Standards alignment

This lesson covers the following Common Core State Standards for Mathematics.

CCSS.MATH.CONTENT.HSG.SRT.D.10Similarity, Right Triangles, and Trigonometry(+) Prove the Laws of Sines and Cosines and use them to solve problems.
CCSS.MATH.CONTENT.HSG.SRT.D.11Similarity, Right Triangles, and Trigonometry(+) Understand and apply the Law of Sines and the Law of Cosines to find unknown measurements in right and non-right triangles (e.g., surveying problems, resultant forces).

For teachersPrint problems onlyTick to print a blank worksheet; leave it clear to print the lesson with the full answer key.

Quick answer

What you'll learn

Triangles without a right angle

Why the ratios match

Two angles and a side

The ambiguous case

Worked examples

Common mistakes

Practice problems

Frequently asked questions

What is the Law of Sines?

When do I use the Law of Sines?

What is the ambiguous case?

How do I prove the Law of Sines?

Does it work for right triangles?

What to learn next

Standards alignment

More Geometry lessons