Law of Cosines

The formula

For any triangle with sides a, b, c and angles A, B, C (where A is opposite side a, etc.):

c² = a² + b² − 2ab cos(C)
b² = a² + c² − 2ac cos(B)
a² = b² + c² − 2bc cos(A)

By symmetry, you can write the formula for any side as a function of the other two and the angle between them. The "−2ab cos(C)" is the correction term that handles non-right angles.

When C = 90°, cos(C) = 0 and you get the Pythagorean theorem — c² = a² + b². When C is acute (less than 90°), cos(C) is positive, so you subtract — c is shorter than the right-triangle hypotenuse would be. When C is obtuse (greater than 90°), cos(C) is negative, so you add (subtracting a negative) — c is longer.

Two main use cases

SAS — Side-Angle-Side

Given two sides and the included angle, find the third side.

Example — a = 5, b = 7, C = 60°. Find c.

c² = 5² + 7² − 2·5·7·cos(60°)
   = 25 + 49 − 70·(0.5)
   = 74 − 35 = 39
c = √39 ≈ 6.24

SSS — Side-Side-Side

Given all three sides, find any angle. Rearrange the formula to solve for cos(C):

cos(C) = (a² + b² − c²) / (2ab)

Example — a = 8, b = 11, c = 13. Find angle C (opposite side c = 13).

cos(C) = (64 + 121 − 169) / (2·8·11)
       = 16 / 176 = 0.0909
C = arccos(0.0909) ≈ 84.8°

If cos(C) comes out negative, the angle is obtuse. If it comes out greater than 1 or less than −1, the three sides don't form a valid triangle (they fail the triangle inequality).

Vector form — same equation, deeper meaning

Let u and v be two vectors with magnitudes a, b and angle θ between them. The vector u − v has magnitude c, the third side of the triangle they form. Then:

|u − v|² = |u|² + |v|² − 2 u · v
         = a² + b² − 2ab cos θ

The dot product u · v = |u||v|cos θ encodes both magnitudes and the angle. The law of cosines is the dot product formula expanded in terms of triangle sides. This is why every linear algebra course derives angles between vectors using essentially the law of cosines.

Proof from coordinates

Place the triangle with one vertex at the origin and another at (b, 0). The third vertex is at (a cos C, a sin C). The third side has length:

c = √((a cos C − b)² + (a sin C − 0)²)
c² = (a cos C − b)² + (a sin C)²
   = a² cos²C − 2ab cos C + b² + a² sin²C
   = a²(cos²C + sin²C) − 2ab cos C + b²
   = a² + b² − 2ab cos C

Using sin²C + cos²C = 1 in the third-to-last step. The law of cosines is just coordinate geometry plus Pythagoras.

Law of cosines vs law of sines

	Law of cosines	Law of sines
Formula	c² = a² + b² − 2ab cos(C)	a/sin A = b/sin B = c/sin C
Solves	SAS, SSS	ASA, AAS, SSA (ambiguous)
Reduces to Pythagoras	Yes (when C = 90°)	No
Ambiguous case	No	SSA can give 0, 1, or 2 solutions
Best for finding	Sides given angle, angles given sides	Sides given two angles + opposite side

Most triangle problems can be solved with either, but one is usually shorter. Memorize when to reach for which.

JavaScript: triangle solver

// Find third side given two sides + included angle (in radians)
function thirdSide(a, b, C) {
  return Math.sqrt(a*a + b*b - 2*a*b*Math.cos(C));
}

// Find angle opposite given side, given all three sides
function angleFromSides(a, b, c) {
  // Returns angle C in radians, opposite side c
  const cosC = (a*a + b*b - c*c) / (2*a*b);
  return Math.acos(cosC);
}

// Validate triangle inequality
function validTriangle(a, b, c) {
  return a + b > c && b + c > a && a + c > b;
}

// Solve for all unknown angles given three sides
function solveSSS(a, b, c) {
  if (!validTriangle(a, b, c)) throw new Error('Invalid triangle sides');
  const A = angleFromSides(b, c, a);
  const B = angleFromSides(a, c, b);
  const C = Math.PI - A - B;  // angles sum to π
  return { A, B, C, degrees: { A: A*180/Math.PI, B: B*180/Math.PI, C: C*180/Math.PI } };
}

solveSSS(8, 11, 13);
// { A: 0.673, B: 0.999, C: 1.470, degrees: { A: 38.6°, B: 57.2°, C: 84.2° } }

// Distance between two GPS points using law of cosines (spherical)
function gpsDistance(lat1, lon1, lat2, lon2) {
  const R = 6371;  // Earth's radius in km
  const φ1 = lat1 * Math.PI/180;
  const φ2 = lat2 * Math.PI/180;
  const Δλ = (lon2 - lon1) * Math.PI/180;
  return R * Math.acos(Math.sin(φ1)*Math.sin(φ2) + Math.cos(φ1)*Math.cos(φ2)*Math.cos(Δλ));
}

More worked examples

Surveyor finding a distance across a lake

Two points A and B on shore are 300m apart. From A, the angle to a third point C (across the lake) is 73°. From B, the angle to C is 58°. Find AC.

Use law of sines first to find AC. The angle at C is 180° − 73° − 58° = 49°. Then:

AC / sin(58°) = 300 / sin(49°)
AC = 300 · sin(58°) / sin(49°) ≈ 300 · 0.848 / 0.755 ≈ 337m

Or, using law of cosines with sides AB = 300 and the two computed sides — verify by computing AC directly. Both approaches agree.

Verify a triangle with sides 7, 24, 25 is right-angled

Compute cos C where C is opposite side 25:

cos(C) = (7² + 24² − 25²) / (2·7·24) = (49 + 576 − 625) / 336 = 0/336 = 0
C = arccos(0) = 90°

It's a right triangle. The 7-24-25 is a Pythagorean triple, confirming the right angle at C.

Where the law of cosines is essential

• Surveying. Triangulation — measure two sides and the included angle, compute the third side. Used in land surveying since at least 1600.
• Navigation and GPS. Ship and aircraft course corrections, distance to a destination given known reference points. Spherical law of cosines for great-circle distances on Earth.
• Computer graphics. Computing angles between vectors (lighting, normals, view direction) — the dot product is the law of cosines under the hood.
• Robotics — inverse kinematics. Given a desired end-effector position and link lengths, compute joint angles. Each joint's angle solves a triangle problem.
• Crystallography. Determining lattice parameters from X-ray diffraction angles. Bragg's law combines with the law of cosines for crystal structure determination.
• Sports analytics. Computing angles for shots in soccer, basketball, golf — given player position, target position, and obstacles, what's the available angle?

Common mistakes

• Using law of cosines on the wrong side. The angle C must be opposite side c (and adjacent to sides a and b). If you mix up which angle is between which sides, you'll plug in wrong values.
• Forgetting the minus sign on cos(C). The formula has −2ab cos(C). Substituting +2ab cos(C) gives wrong answers and is geometrically nonsensical (would make c larger when C is acute).
• Confusing degrees and radians. Math.cos in JavaScript expects radians. cos(60°) is not what you want; cos(60 * π/180) = cos(π/3) = 0.5 is.
• Triangle inequality violated. If you compute sides that don't satisfy a + b > c (or any cyclic permutation), no real triangle exists. arccos returns NaN. Validate inputs before computing.
• Using law of cosines for SSA. SSA (two sides + non-included angle) is the ambiguous case best handled by law of sines (which detects 0, 1, 2 solutions). Forcing law of cosines gives a quadratic — solve it carefully.
• Wrong cos sign for obtuse angles. cos is negative for angles between 90° and 180°. Forgetting this when C is obtuse — c² = a² + b² + 2ab|cos C| (note + because cos is negative — the minus and the negative cancel). Just trust the formula; the algebra handles signs.