DWG NO. 6.2 — Lesson 2 of 6

Law of Cosines

Unit 6: Additional Topics in Trigonometry · ~25 min

Objective Use the Law of Cosines to solve oblique triangles given SAS or SSS, and to find any angle once all three sides are known.

The Law of Sines needs a known angle paired with its opposite side. Two common triangle setups never hand you that pair: SAS (two sides and the angle between them) and SSS (all three sides, no angles at all). For these, the Law of Cosines — a direct generalization of the Pythagorean theorem — is the tool.

The Law of Cosines

Using the same labeling as Lesson 6.1 (side a opposite angle A, and so on):

c² = a² + b² − 2ab·cosC
a² = b² + c² − 2bc·cosA
b² = a² + c² − 2ac·cosB

Each version pairs one side with the angle directly across from it, using the other two sides and the angle between them. Notice that if C = 90°, cosC = 0 and the formula collapses to c² = a² + b² — the Pythagorean theorem is just the special case where the included angle is a right angle.

Side a is opposite angle A, which sits between the known sides b and c — the SAS setup the Law of Cosines solves directly

Solving SAS triangles

Given two sides and the included angle, compute the third side first with the Law of Cosines, then find a remaining angle (with either the Law of Cosines or the Law of Sines) and subtract from 180° for the last one.

Solving SSS triangles

Given all three sides, solve the Law of Cosines for cosine and find any angle directly — there's no ambiguous case here, since inverse cosine over 0° to 180° gives exactly one angle for each ratio:

cosC = \(\frac{a^{2} + b^{2} - c^{2}}{2ab}\)

Worked Example 1 · SAS case

Problema = 8, b = 10, C = 53°. Find c, then angle A.

1c² = 8² + 10² − 2(8)(10)cos53° = 164 − 160(0.6018) ≈ 67.71, so c ≈ 8.23.

2cosA = \(\frac{b^{2} + c^{2} - a^{2}}{2bc}\) = \(\frac{100 + 67.71 - 64}{2\cdot10\cdot8.23}\) ≈ 0.6301, so A ≈ 50.9°.

3B = 180° − 53° − 50.9° ≈ 76.1°.

c ≈ 8.23, A ≈ 50.9°, B ≈ 76.1°

Worked Example 2 · SSS case

Problema = 7, b = 9, c = 13. Find angle C, the largest angle (opposite the longest side).

1cosC = \(\frac{a^{2} + b^{2} - c^{2}}{2ab}\) = \(\frac{49 + 81 - 169}{2\cdot7\cdot9}\) = \(\frac{-39}{126}\) ≈ −0.3095.

2C ≈ 108.0°. The negative cosine confirms C is obtuse, which makes sense since c is the longest side.

C ≈ 108.0°

Guided practice

1.SAS: a = 6, b = 9, C = 40°. Find c.
Show answer

c² = 36 + 81 − 2(6)(9)cos40° = 117 − 108(0.7660) ≈ 34.27, so c ≈ 5.85.
2.SSS: a = 5, b = 6, c = 7. Find angle C.
Show answer

cosC = \(\frac{25 + 36 - 49}{60}\) = \(\frac{12}{60}\) = 0.2, so C ≈ 78.5°.
3.SSS: a = 10, b = 10, c = 15. Find the angle opposite c.
Show answer

cosC = \(\frac{100 + 100 - 225}{200}\) = \(\frac{-25}{200}\) = −0.125, so C ≈ 97.2°.
4.Explain why the Law of Cosines has no ambiguous case, unlike SSA under the Law of Sines.
Show answer

Given two sides and the included angle, c² = a² + b² − 2ab·cosC produces exactly one positive value of c. And when solving for an angle from cosC = …, the inverse cosine function is one-to-one over 0°–180°, so each ratio gives exactly one angle — unlike inverse sine, which can't distinguish an angle from its supplement.
5.SAS: a = 13, c = 20, B = 110° (angle between sides a and c). Find b.
Show answer

b² = 13² + 20² − 2(13)(20)cos110° = 569 − 520(−0.3420) ≈ 746.84, so b ≈ 27.3.

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