Chapter 6.2, Problem 3E

To determine

To complete the statement.

Answer to Problem 3E

When solving an oblique triangle given two sides and their included angle, use the standard form of the

Law of Cosines to solve for remaining side.

Explanation of Solution

Given information:

When solving an oblique triangle given two sides and their included angle, use the $\underset{\_}{}$ form of the

Law of Cosines to solve for remaining side.

Calculation:

Let ABC is an oblique triangle,

  

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

Let sides are $a=3,c=4,\text{angle}B=60°$ .

Substitute these values in the Law of cosine formula,

  $\begin{array}{l}{b}^2=a^2+c^2-2a\cdot c\cdot \cos B\\ b^2=3^2+4^2-2\cdot 3\cdot 4\cdot \cos 60°\\ b^2=9+16-24\cdot \left(\frac{1}{2}\right)\\ b^2=25-12\\ b^2=13\end{array}$

Take square root on both sides.

  $\begin{array}{c}\sqrt{b^2}=\sqrt{13}\\ b=3.6055\end{array}$

So, the missing side is $b=3.6055$ .

So, when solving an oblique triangle given two sides and their included angle, use the standard form of the

Law of Cosines to solve for remaining side.