How do I use the law of cosines?

How it works

For a triangle with sides a, b, c and the angle C between sides a and b (opposite side c), the law of cosines says:

c² = a² + b² − 2ab cos C

Fill in any three of a, b, c and C, and the calculator finds the fourth.

• Side c (SAS): c = √(a² + b² − 2ab cos C). The calculator uses the equal form √((a − b)² + 4ab sin²(C/2)), which keeps its accuracy for small angles.
• Angle C (SSS): C = arccos((a² + b² − c²) ÷ 2ab). The three sides must form a triangle: each side shorter than the other two together.
• Side a or b (SSA): a is a root of a² − 2b cos C · a + (b² − c²) = 0, so a = b cos C ± √(c² − b² sin² C). Only positive roots count. There can be no triangle (the square root of a negative number), one, or two (the ambiguous case), and the calculator shows every one.

Once the triangle is known, the calculator works out the rest from the three values you typed, in forms that stay accurate for needle-like triangles (one very short side or one very small angle; W. Kahan, "Miscalculating Area and Angles of a Needle-like Triangle", 2014):

• Angles A and B, with angle C and two sides typed (SAS, or SSA with side a or b solved). From the law of sines and a projection, c sin A = a sin C and c cos A = b − a cos C, so A = atan2(a sin C, b − a cos C), and likewise B = atan2(b sin C, a − b cos C). When side a was solved from b, c and C, a − b cos C = ±√(c² − b² sin² C) (+ for the larger root, − for the smaller), and b − a cos C = b sin² C − cos C × (a − b cos C); the same holds with a and b swapped. Both angles are computed on their own, not as 180° minus the other two, because a small angle loses its digits in that subtraction.
• Angles with three sides typed (SSS). Each angle from Kahan's formula: for the angle X opposite side x, with the other sides p ≥ q, X = 2 arctan √(((p − q) + x) × μ ÷ ((p + (q + x)) × ((p − x) + q))), where μ = x − (p − q) when q ≥ x, and μ = q − (p − x) otherwise. Angle C itself is found the same way.
• Perimeter P = a + b + c, and the semi-perimeter s = P ÷ 2.
• Area: ½ab sin C when angle C was typed; Heron's formula √(s(s − a)(s − b)(s − c)) in Kahan's rearranged form when the three sides were typed.
• Inradius r = area ÷ s, and circumradius R = c ÷ (2 sin C). In the ambiguous case both triangles share c and C, so the circumradius is shown once.

Assumptions

• Sides are positive numbers, all in the same unit. Angle C is more than 0° and less than 180°.
• Angle C can be typed in degrees or radians. Angles A and B are shown in degrees.
• A circumradius or other value too large for a 64-bit float (for an angle C very close to 0) is left out instead of shown as infinity.
• An old link with goal=angle leaves angle C empty, so it is solved from the three sides. An old link with useRadians=true reads angle C in radians.
• When three sides are given that make no triangle (one side is as long as the other two together, or longer), the page gives no answer and says: "These sides do not make a triangle: each side must be shorter than the other two together."

Worked examples by hand

a = 5, b = 7, C = 60° (SAS). c² = 25 + 49 − 2 × 5 × 7 × cos 60° = 74 − 70 × 0.5 = 39, so c = √39 = 6.244998. The perimeter is 5 + 7 + 6.244998 = 18.244998, and the area is ½ × 5 × 7 × sin 60° = 15.155445.

a = 3, b = 4, C = 90°. c² = 9 + 16 − 0 = 25, so c = 5. Angle A = arcsin(3/5) = 36.87° (0.6435 rad). The area is ½ × 3 × 4 = 6, the inradius 6 ÷ 6 = 1, and the circumradius 3 × 4 × 5 ÷ 24 = 2.5.

a = 5, b = 7, c = 8 (SSS). cos C = (25 + 49 − 64) ÷ 70 = 1/7, so C = arccos(1/7) = 81.79° (1.4274 rad). cos B = (25 + 64 − 49) ÷ 80 = 1/2, so B = 60°. By Heron's formula with s = 10, the area is √(10 × 5 × 3 × 2) = √300 = 17.320508.

b = 8, c = 7, C = 60° (SSA). a² − 2 × 8 × 0.5 × a + (64 − 49) = 0, so a² − 8a + 15 = 0 and a = 5 or 3. Both triangles fit.

a = 5, b = 7, C = 2 radians. c² = 74 − 70 × cos 2 = 74 + 29.130 = 103.130, so c = 10.155308.