Law of Cosines

From the first congruence criterion for triangles, two sides and the included angle uniquely determine a triangle. The Law of Cosines tells us how to find the third side.

⚡Theorem — Law of Cosines (Theorem 3.1)

The square of any side of a triangle equals the sum of the squares of the other two sides minus twice their product times the cosine of the included angle:

$\boxed{a^2 = b^2 + c^2 - 2bc\cos\alpha}$

where $a$ is the side opposite angle $\alpha$ , and $b$ , $c$ are the other two sides.

Note: When $\alpha = 90°$ , $\cos\alpha = 0$ and the formula reduces to the Pythagorean theorem $a^2 = b^2 + c^2$ . This shows the Pythagorean theorem is a special case of the Law of Cosines.

Adjust the sides and angle to see how the formula $a^2 = b^2 + c^2 - 2bc\cos\alpha$ responds. Notice how the correction term changes sign as $\alpha$ crosses $90°$ :

b = 3.0

c = 4.0

α = 60.0°

a² = b²+c²-12.00 = 13.000a = 3.606α = 60°β = 46.1°γ = 73.9°▲ acute

Determining Triangle Type

⚡Theorem — Corollary (Theorem 3.2)

Let $a$ , $b$ , $c$ be the sides of a triangle where $a$ is the longest side. Then:

If $a^2 < b^2 + c^2$ , the triangle is acute.
If $a^2 > b^2 + c^2$ , the triangle is obtuse.
If $a^2 = b^2 + c^2$ , the triangle is right-angled.

✎Example — Finding a Side

In triangle $ABC$ : $AB = 14$ cm, $BC = 13$ cm, $AC = 15$ cm. A point $D$ on $AC$ satisfies $CD : AD = 1 : 2$ . Find $BD$ .

Solution. First find $\cos C$ using the Law of Cosines from triangle $ABC$ :

$AB^2 = AC^2 + BC^2 - 2 \cdot AC \cdot BC \cdot \cos C$

$\cos C = \frac{AC^2 + BC^2 - AB^2}{2 \cdot AC \cdot BC} = \frac{225 + 169 - 196}{2 \cdot 15 \cdot 13} = \frac{198}{390} = \frac{33}{65}$

Since $CD : AD = 1:2$ , we have $CD = \tfrac{1}{3}AC = 5$ cm. From triangle $BCD$ :

$BD^2 = BC^2 + CD^2 - 2 \cdot BC \cdot CD \cdot \cos C = 169 + 25 - 2 \cdot 13 \cdot 5 \cdot \frac{33}{65} = 128$

$BD = 8\sqrt{2} \text{ cm} \qquad \blacktriangleleft$

Parallelogram Diagonal Theorem

⚡Theorem — Theorem 3.3

The sum of the squares of the diagonals of a parallelogram equals the sum of the squares of all four sides:

$d_1^2 + d_2^2 = 2(a^2 + b^2)$

Median Formulas

As a consequence of the Law of Cosines, the lengths of the medians of a triangle with sides $a$ , $b$ , $c$ can be expressed as:

$m_a^2 = \frac{2b^2 + 2c^2 - a^2}{4}, \quad m_b^2 = \frac{2c^2 + 2a^2 - b^2}{4}, \quad m_c^2 = \frac{2a^2 + 2b^2 - c^2}{4}$

✎Example — Checking Triangle Type

The sides of a triangle are $\sqrt{18}$ cm, $5$ cm, and $7$ cm. Determine the triangle type.

Solution. The longest side is $7$ cm. Check: $7^2 = 49$ vs $(\sqrt{18})^2 + 5^2 = 18 + 25 = 43$ .

Since $49 > 43$ , the triangle is obtuse. $\blacktriangleleft$