Chapter 6.6, Problem 1E

For triangle ABC with sides a, b, and c the Law of Cosines states

$c^2=\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$

To determine

To fill: The Law of Cosines for triangle $ABC$ with sides $a$, $b$ and $c$.

Answer to Problem 1E

The Law of Cosines states $c^2=a^2+b^2-2ab\cos C$ for triangle $ABC$ with sides $a$, $b$ and $c$.

Explanation of Solution

Given:

The triangle $ABC$ with sides $a$, $b$ and $c$.

Calculation:

Place triangle ABC so that $\angle C$ is at the origin.

Figure (1)

From Figure (1), it is noticed that the coordinates of vertices $A$ and $B$ are $\left(b,0\right)$ and $\left(a\cos C,a\sin C\right)$ respectively. Using the Distance Formula,

$\begin{array}{c}{c}^2={\left(a\cos C-b\right)}^2+{\left(a\sin C-0\right)}^2\\ =a^2{\cos }^{2}C-2ab\cos C+b^2+a^2{\sin }^{2}C\\ =a^2\left({\cos }^{2}C+{\sin }^{2}C\right)-2ab\cos C+b^2\\ =a^2+b^2-2ab\cos C\left(\because {\sin }^{2}C+{\cos }^{2}C=1\right)\end{array}$

Thus, the Law of Cosines states $c^2=a^2+b^2-2ab\cos C$ for triangle $ABC$ with sides $a$, $b$ and $c$.