Section 10.2 - Law of Cosines

Section Objectives

1. Use the Law of Cosines to solve triangles.
2. Use the Law of Cosines in applications.
3. Use Heron's formula to find the area of a triangle.

Law of Cosines

Suppose $ABC$ is a triangle with sides of lengths $a$, $b$, and $c$ that are opposite their respective vertices $A$, $B$, and $C$. Let $\alpha$, $\beta$, and $\gamma$ be the angles at $A$, $B$, and $C$. The Law of Cosines says that

$a^2=b^2+c^2-2bc \,\cos \alpha, \quad b^2 = a^2 + c^2 - 2ac \, \cos \beta, \quad c^2=a^2+b^2 -2ab \, \cos \gamma$.

When the appropriate information is known, the Law of Cosines can be used to solve triangles.

Examples

• $a=8$ feet, $b=19$ feet, $c=14$ feet. Solve the triangle. (This is an SSS case.)

• $b=9$ meters, $c=12$ meters, $\alpha = 25^{\circ}$. Solve the triangle. (This is an SAS case.)

Heron's Area Formula

Heron's formula for the area of $\triangle ABC$ can be derived from the Law of Cosines:

Area $= \sqrt{s(s-a)(s-b)(s-c)}$, where $s$ is the semi-perimeter of the triangle with sides of lengths $a$, $b$, and $c$.

Examples

• Find the area of a triangular in the first example above.