Law of Cosines
Derive and apply the Law of Cosines (c² = a² + b² − 2ab·cosC) to solve triangles given SAS or SSS. Recognize it as a generalization of the Pythagorean theorem and determine when to use Sines vs. Cosines.
A lesson of Geometry, in Right Triangles and Trigonometry, about 10 minutes.
By the end of this lesson, you will be able to state and apply the Law of Cosines to solve SSS and SAS triangles, determine when to use the Law of Cosines versus the Law of Sines, and verify that the Pythagorean theorem is a special case.
The Big Idea
The Law of Sines requires a matched angle-side pair. When you have SAS (two sides and the included angle) or SSS (all three sides but no angles), there's no matched pair to start — the Law of Sines cannot begin. The Law of Cosines handles exactly these cases, and it generalizes the Pythagorean theorem to non-right triangles.
The Law of Cosines
For $\triangle ABC$ with sides $a, b, c$ opposite angles $A, B, C$:
$$a^2 = b^2 + c^2 - 2bc\cos A$$ $$b^2 = a^2 + c^2 - 2ac\cos B$$ $$c^2 = a^2 + b^2 - 2ab\cos C$$
Pythagorean theorem as a special case: when $C = 90°$, $\cos C = 0$, so $c^2 = a^2 + b^2 + 0 = a^2 + b^2$. ✓
Solving for an Angle
Rearrange to find an angle from three sides:
$$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$$
When to Use Which Law
| Given | Use |
|---|---|
| AAS or ASA (angle-side pair available) | Law of Sines |
| SAS (two sides + included angle) | Law of Cosines |
| SSS (all three sides) | Law of Cosines |
| SSA (ambiguous case) | Law of Sines (with caution) |
Strategy: use Law of Cosines first when no complete angle-side pair exists. After finding one more element, you can often switch to the simpler Law of Sines.
Example 1 — SAS
$\triangle ABC$: $b = 10$, $c = 14$, $A = 48°$. Find $a$.
$$a^2 = 100 + 196 - 2(10)(14)\cos 48° = 296 - 280(0.6691) = 296 - 187.3 \approx 108.7$$
$$a \approx \sqrt{108.7} \approx 10.43$$
Now use Law of Sines to find another angle:
$$\frac{\sin B}{10} = \frac{\sin 48°}{10.43} \implies \sin B \approx \frac{10(0.7431)}{10.43} \approx 0.7125 \implies B \approx 45.4°$$
$C = 180 - 48 - 45.4 = 86.6°$
Example 2 — SSS
$\triangle ABC$: $a = 7$, $b = 9$, $c = 11$. Find all angles.
$$\cos A = \frac{81 + 121 - 49}{2(9)(11)} = \frac{153}{198} \approx 0.7727 \implies A \approx 39.5°$$
$$\cos B = \frac{49 + 121 - 81}{2(7)(11)} = \frac{89}{154} \approx 0.5779 \implies B \approx 54.7°$$
$C = 180 - 39.5 - 54.7 = 85.8°$
Verify: $\cos C = \frac{49 + 81 - 121}{2(7)(9)} = \frac{9}{126} \approx 0.0714 \implies C \approx 85.9°$ ✓