What is the law of sines?

In any triangle the ratio of a side to the sine of the angle opposite it is the same for all three pairs: a/sin α = b/sin β = c/sin γ. That common ratio equals 2R, where R is the radius of the circle through the three vertices.

What is the law of cosines?

a² = b² + c² − 2bc·cos α, where α is the angle enclosed between the sides b and c. The analogous formulas hold for the other sides — you just rotate the letters.

When do you use the law of sines and when the law of cosines?

Use the law of sines when you know a side together with the angle opposite it (two angles and a side, or two sides and an angle opposite one of them). Use the law of cosines when you know two sides and the angle between them, or all three sides.

Why is the law of cosines called a generalisation of the Pythagorean theorem?

Because for α = 90° we have cos 90° = 0, so the term −2bc·cos α vanishes and a² = b² + c² is left. The Pythagorean theorem is thus a special case of the law of cosines, and the cosine term is the correction for how far the angle is from a right one.

How do you find an angle from three sides?

Rearrange the law of cosines: cos α = (b² + c² − a²) / (2bc), then read the angle off with the inverse function (arccos). A negative result simply means the angle is obtuse — that is normal, not an error.

How do you find the area of a triangle from two sides and the angle between them?

From P = ½ab·sin γ, where γ is the angle contained between the sides a and b. For a = 8 cm, b = 6 cm and γ = 30° the area is ½ · 8 · 6 · 0.5 = 12 cm². The formula generalises the school rule "half the base times the height" — the altitude onto side a is exactly b·sin γ long.

Does P = ½ab·sin γ work for an obtuse angle?

Yes, with no correction at all. The sine of an obtuse angle is positive, so the area comes out positive — unlike the law of cosines, where the cosine changes sign. Moreover sin γ = sin(180° − γ), so a triangle with a 150° angle between the same two sides has exactly the same area as one with 30°.

Can the law of sines give two answers?

Yes, in the case of two sides and an angle opposite the shorter of them (the SSA case). Since sin α = sin(180° − α), the same sine value describes both an acute and an obtuse angle. Check both and discard the one whose angle sum would exceed 180°.

Do these laws work in a right triangle?

Yes, both hold in every triangle. In a right one it is usually quicker to use the definitions of sine and cosine or the Pythagorean theorem — they give the same answer with less arithmetic.

Where are these laws used in practice?

In triangulation: surveying, navigation and field measurement determine a distance nobody can measure directly by measuring a baseline and two angles. The height of a mountain, the width of a river and the distance of a ship from shore are all computed this way.

Advanced level

The laws of sines and cosines

The law of sines and the law of cosines carry trigonometry into any triangle — including one with no right angle. See which to pick for the data you are given, why the law of cosines generalises Pythagoras, and when there are two solutions.

Before you start

This topic builds on earlier ideas. Before you start, it's worth working through the lessons below — they'll make everything click:

Sine, cosine and tangentSine, cosine and tangent are three ratios of the sides of a right triangle. See why they depend on the angle alone, where the exact values for 30°, 45° and 60° come from, and how to find a side you cannot measure.
AnglesAngles are measured in degrees: acute below 90°, right exactly 90°, obtuse above it. Meet the kinds of angle, complementary and supplementary pairs, the alternate and corresponding angles at two parallel lines, and the two sum rules — 180° in a triangle and (n − 2) · 180° in a polygon.
The unit circleThe unit circle carries the sine and the cosine from a triangle onto the whole circle. See how the coordinates of a point become the values of the functions, what signs they take in the four quadrants, and what a negative angle or one past a full turn means.

Where this is used

Real situations where you count exactly the way this lesson teaches:

Fencing a plot
Two sides of a plot run 12 m and 9 m and meet at 115°. The law of cosines closes the corner: √(12² + 9² − 2 · 12 · 9 · cos 115°) = √(225 + 91.3) = 17.8 m. The cosine of an obtuse angle is negative, so the subtraction turns into an addition — 2.8 m more fencing than the 15 m a square corner would have needed.
Field surveying
You do not measure a river with a tape. A surveyor lays out a 50 m baseline along the bank and, from its two ends, reads the angles to a tree on the far side: 78° and 65°. The third angle is 180° − 78° − 65° = 37°, and the law of sines gives the distance from the 78° end to the tree: 50 · sin 65° / sin 37° = 75.3 m. That is not yet the width — the width is the tree’s distance from the baseline, 75.3 · sin 78° = 73.6 m. One baseline and two angles also settle the height of a chimney or a ship’s distance off the quay.
Valuing a plot
A triangular plot has two sides of 42 m and 35 m meeting at 68°. Its area is ½ · 42 · 35 · sin 68° = 735 · 0.927 = 681 m² — without walking the ground and without measuring a height that could not be staked out through undergrowth anyway. At 180 zł a square metre that values the plot at 122,600 zł, and the difference between 68° and a carelessly read 65° is already 27 m², close to 5,000 zł.
Hiking
You follow a path for 3 km, turn through 120° and walk 2 km more. The start lies √(3² + 2² − 2 · 3 · 2 · cos 60°) = √7 ≈ 2.6 km away in a straight line, though your legs covered 5 km. Mind the angle: the sum takes 60°, because a turn is measured off your heading while the law of cosines wants the angle between the legs — 120° would say 4.4 km and argue for a different way home.

All formulas

The law of sines
$\frac{a}{\sin \alpha} = \frac{b}{\sin \beta} = \frac{c}{\sin \gamma}$
a side and the angle opposite it — always as a pair
A side from the law of sines
$b = \frac{a \cdot \sin \beta}{\sin \alpha}$
the proportion solved for the unknown side
The law of cosines
$a^2 = b^2 + c^2 - 2bc \cos \alpha$
α is the angle between the sides b and c
An angle from three sides
$\cos \alpha = \frac{b^2 + c^2 - a^2}{2bc}$
the law of cosines solved for the angle
The angle sum
$\alpha + \beta + \gamma = 180^\circ$
the third angle always follows from the other two
Area of a triangle
$P = \frac{1}{2} ab \sin \gamma$
two sides and the angle BETWEEN them — no height needed

Sine, cosine and tangent were defined in a right triangle. Most triangles, in exercises and in the field, have no right angle — and yet a missing side and a missing angle can still be computed. Two theorems do the job.

Labelling

Side a lies opposite angle α, side b opposite β, side c opposite γ. That 'side and the angle opposite it' pairing is the key to both laws.

The convention is simple: a lowercase letter is the side opposite the angle carrying the matching Greek letter. On top of that, the angle sum is always available:

\alpha + \beta + \gamma = 180^\circ

so two known angles hand you the third for free.

The law of sines

\frac{a}{\sin \alpha} = \frac{b}{\sin \beta} = \frac{c}{\sin \gamma}

A side and the sine of the angle opposite it are proportional: a longer side lies opposite a larger angle. The common ratio even has a geometric meaning — it equals $2R$ , where $R$ is the radius of the circle through the three vertices.

Use the law when the data contains a complete pair: a side together with the angle opposite it.

b = \frac{a \cdot \sin \beta}{\sin \alpha}

In a triangle a = 10 cm, α = 40° and β = 65°. Find side b (to 0.1 cm).

The law of cosines

a^2 = b^2 + c^2 - 2bc \cos \alpha

Here $\alpha$ is the angle enclosed between the sides $b$ and $c$ . Use the law when there is no complete side–angle pair: two sides and the angle between them, or all three sides.

Look at how the formula is built: the first two terms are the Pythagorean theorem, and the third is a correction. For $\alpha = 90^\circ$ we have $\cos 90^\circ = 0$ , so the correction disappears:

a^2 = b^2 + c^2

The Pythagorean theorem is therefore a special case of the law of cosines. For an acute angle the cosine is positive and the correction shrinks $a^2$ ; for an obtuse angle the cosine is negative, minus times minus makes plus, and the opposite side grows longer — exactly what the picture suggests.

In a triangle b = 3 cm, c = 8 cm and the angle between them is α = 60°. Find side a.

A triangle has sides 5 cm, 7 cm and 8 cm. Is it acute?

The area from two sides and the angle

The very data the law of cosines needs — two sides and the angle between them — is also enough to find the area, and without working out the third side first.

The starting point is the school formula $P = \frac{1}{2} a h$ , where $h$ is the altitude onto side $a$ . The trouble is that in a general triangle nobody hands you an altitude. But it can be computed.

The altitude dropped from vertex A onto side a is a leg of a right triangle with hypotenuse b and angle γ.

The altitude, the side $b$ and part of side $a$ form a right triangle. In it $b$ is the hypotenuse and $h$ the leg opposite the angle $\gamma$ , so straight from the definition of the sine:

\sin \gamma = \frac{h}{b} \quad \Longrightarrow \quad h = b \sin \gamma

Substituting into the area formula:

P = \frac{1}{2} a h = \frac{1}{2} ab \sin \gamma

The letters can be permuted — all that matters is that the angle sits between the two sides being multiplied:

P = \frac{1}{2} ab \sin \gamma = \frac{1}{2} bc \sin \alpha = \frac{1}{2} ac \sin \beta

Two checks. For $\gamma = 90^\circ$ we get $\sin 90^\circ = 1$ and what remains is $P = \frac{1}{2} ab$ , half the product of the legs ✓. And since $\sin \gamma = \sin(180^\circ - \gamma)$ , a triangle with $150^\circ$ between the same two sides has the same area as one with $30^\circ$ — the formula needs no separate obtuse case, unlike the law of cosines.

In a triangle a = 8 cm, b = 6 cm and the angle between them is γ = 30°. Find the area.

Does a triangle with sides 10 cm and 7 cm and a 140° angle between them have a larger area than the same triangle with 40°?

Choosing the law

given	law
two angles and a side	sines (third angle from the $180^\circ$ sum)
two sides and an angle opposite one of them	sines (careful: two solutions are possible)
two sides and the angle between them	cosines
three sides	cosines (solved for the angle)
two sides and the angle between them, but the area is wanted	the formula $P = \frac{1}{2} ab \sin \gamma$

The rule in one sentence: a complete side–opposite-angle pair → sines; no such pair → cosines. And when the question is about area rather than a side, the sine formula settles it outright.

The ambiguous case

Data of the form "two sides and the angle opposite the shorter one" can be ambiguous. The reason is in the function itself: $\sin \alpha = \sin(180^\circ - \alpha)$ , so one value of the sine belongs to two angles — one acute, one obtuse.

Take $a = 6$ , $b = 8$ and $\alpha = 40^\circ$ . The law of sines gives

\sin \beta = \frac{b \sin \alpha}{a} = \frac{8 \cdot 0.6428}{6} \approx 0.857

and that sine belongs to both $\beta \approx 59^\circ$ and $\beta \approx 121^\circ$ . Both are admissible, because in both cases the angle sum stays below $180^\circ$ — so two different triangles fit the data. You only discard the one where $\alpha + \beta$ would exceed $180^\circ$ .

A calculator will not point this out: the inverse function returns the acute angle only. The second solution has to be added by hand.

The calculator: DEG mode

Both laws are computed with angles in degrees, so the calculator has to sit in DEG mode. The quick test: $\sin 30$ must give $0.5$ . If it gives $-0.988$ , the calculator is working in radians and every result will be wrong even though the algebra is right.

Exercises

Law-of-cosines questions have a whole-number answer — type it together with its unit, e.g. 7 cm. In law-of-sines questions the unit is given in brackets, so type the bare number rounded to 0.1. Area questions follow the same rule: when no unit appears in the prompt, write it into the answer (e.g. 12 cm²); when it is given in brackets, type the bare number rounded to 0.1.

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0

Side a by the law of cosines: b = 8 cm, c = 5 cm, α = 60°

Common mistakes

Pairing a side with the wrong angle — in the law of sines a side always goes with the angle opposite it.
Using the law of cosines with an angle that is not between the given sides — in the formula $\alpha$ is the angle enclosed by $b$ and $c$ .
Computing $2bc \cos \alpha$ as $(2bc \cos \alpha)^2$ — only the sides $b$ and $c$ are squared.
Forgetting the square root — $a^2 = 49$ is not the answer yet; the answer is $a = 7$ .
Missing the second solution in the SSA case — the sine cannot tell an acute angle from an obtuse one.
Computing the area from an angle that is not between the given sides — in $P = \frac{1}{2} ab \sin \gamma$ the angle $\gamma$ has to be enclosed by $a$ and $b$ ; any other angle gives a meaningless result.
Losing the half in the area formula — $ab \sin \gamma$ is the area of the parallelogram on those sides, and a triangle is half of it.
A calculator left in RAD mode — right formulas, wrong numbers.

Formula card

Topic: The laws of sines and cosines

The law of sines
$\frac{a}{\sin \alpha} = \frac{b}{\sin \beta} = \frac{c}{\sin \gamma}$
a side and the angle opposite it — always as a pair
A side from the law of sines
$b = \frac{a \cdot \sin \beta}{\sin \alpha}$
the proportion solved for the unknown side
The law of cosines
$a^2 = b^2 + c^2 - 2bc \cos \alpha$
α is the angle between the sides b and c
An angle from three sides
$\cos \alpha = \frac{b^2 + c^2 - a^2}{2bc}$
the law of cosines solved for the angle
The angle sum
$\alpha + \beta + \gamma = 180^\circ$
the third angle always follows from the other two
Area of a triangle
$P = \frac{1}{2} ab \sin \gamma$
two sides and the angle BETWEEN them — no height needed

Labelling in any triangle: side a lies opposite angle α, side b opposite β, side c opposite γ. Both laws rest on that "side opposite angle" pairing.

The area from two sides and the angle between them: the altitude from A onto side a is b · sin γ long, so P = ½ · a · b · sin γ.

Frequently asked questions

A luminous unit circle with a radius sweeping out an angle, a sine wave unrolling from it as a streak of light, beside a right triangle drawn in stars, against a cosmic nebula in violet and magenta

Trigonometry

Why sine doesn't depend on the size of the triangle

The height of a tower, the width of a river, the distance to a ship offshore — all of it can be computed without going anywhere near the thing being measured. The whole of trigonometry rests on one observation: the ratio of two sides in a right triangle depends only on the angle, not on how big the triangle is. That is what turns an angle into a number — and lets it be tabulated once and for all.

13 min