The Law of Cosines

The law of cosines describes the relationship between the length of the sides of a triangle and the cosine of the angle formed by the triangle. It is often referred to as the cosine rule.

Related Topics

• The Law of Sines
• How to Find Missing Sides and Angles of a Right Triangle
• How to Evaluate Trigonometric Function
• How to Solve Angles and Angle Measure

A step-by-step guide to the law of cosines

If the angles of an oblique triangle are measured by $A$, $B$, and $C$ and $a$, $b$, and $c$ are the lengths of the sides opposite the corresponding angles, then the square of one side of a triangle is equal to the total of the squares of the other two sides minus twice the product of the two sides plus the cosine of the included angle.

$a^2=b^2+c^2-2bc .\cos A$

$b^2=a^2+c^2-2ac .\cos B$

$c^2=a^2+b^2-2ab .\cos C$

Solving for the cosines gives the equivalent formulas:

$cos A=\frac {b^2+c^2-a^2}{2bc}$

$cos B=\frac {c^2+a^2-b^2}{2ca}$

$cos C=\frac {a^2+b^2-c^2}{2ab}$

The Law of Cosines – Example 1:

In the $ABC$ triangle, find the remaining side.

To find side $c$ use the law of cosines: $c^2=a^2+b^2-2ab .\cos C$

$a=14, b=5, C=20$

$c^2$$=14^2+5^2-2(14)(5)(cos 20)=(196+25)-(140×cos 20)=221-(140×0.94)=221-131.6=89.4$

$c^2=89.4$ → $c=\sqrt{89.44}= 9.45$

The Law of Cosines – Example 2:

Find the angle $B$ in the $ABC$ triangle.

To find angle $B$ use the law of cosines: $cos B=\frac {c^2+a^2-b^2}{2ca}$

$b=20, a=8, c=14$

$cos B= \frac {14^2+8^2-20^2}{2(14)(8)} =\frac {196+ 64 – 400}{176}=\frac{-140}{224}=-0.625$

Since $cosB$ is negative, $B$ is an obtuse angle.

$B≅128.69 ^\circ$

Exercises for the Law of Cosines

In the ABC triangle, find the side of c.

1.

2.

3.

1. $\color{blue}{31.12}$
2. $\color{blue}{44.68}$
3. $\color{blue}{21.49}$