Chapter 6.2, Problem 1E

To determine

To complete the statement.

Answer to Problem 1E

The standard form of the Law of cosine for $\cos B=\frac{a^2+c^2-b^2}{2ac}$ is $b^2=a^2+c^2-2a\cdot c\cdot \cos B$ .

Explanation of Solution

Given information:

The standard form of the Law of cosine for $\cos B=\frac{a^2+c^2-b^2}{2ac}$ is $\underset{\_}{}$ .

Calculation:

  $\cos B=\frac{a^2+c^2-b^2}{2ac}$

Cross multiply,

  $2ac\cos B=a^2+c^2-b^2$

Add $b^2$ from both sides,

  $\begin{array}{l}{b}^2+2a\cdot c\cdot \cos B=a^2+c^2-b^2+b^2\\ b^2+2a\cdot c\cdot \cos B=a^2+c^2\end{array}$

Subtract $2a\cdot c\cdot \cos B$ from both sides,

  $\begin{array}{c}{b}^2+2a\cdot c\cdot \cos B-2a\cdot c\cdot \cos B=a^2+c^2-2a\cdot c\cdot \cos B\\ b^2=a^2+c^2-2a\cdot c\cdot \cos B\end{array}$

So, the standard form of the Law of cosine is $b^2=a^2+c^2-2a\cdot c\cdot \cos B$ .