Chapter 6.2, Problem 11E

To determine

To find: The solution for the triangle by law of cosines.

Answer to Problem 11E

The angles of the triangle area $A=30.12°$ , $B=43.16°$ and $C=106.72°$ .

Explanation of Solution

Given:

The sides of the triangle are,

  $\begin{array}{l}a=11\\ b=15\\ c=21\end{array}$

Calculation:

Consider the alternative form of Law of cosine is,

  $\cos C=\frac{a^2+b^2-c^2}{2ab}$

Then,

  $\begin{array}{l}\cos C=\frac{{\left(11\right)}^2+{\left(15\right)}^2-{\left(21\right)}^2}{2\left(11\right)\left(15\right)}\\ \cos C=-0.2878\\ C=106.72°\end{array}$

Consider the law of sine as,

  $\sin B=b\left(\frac{\sin C}{c}\right)$

Then,

  $\begin{array}{l}\sin B=\left(15\right)\left(\frac{\sin 106.72°}{21}\right)\\ \sin B=0.6841\\ B=43.16°\end{array}$

Consider the angle property of the triangle for the third angle of the triangle,

  $A=180°-B-C$

Then,

  $\begin{array}{c}A=180°-43.16°-106.72°\\ =30.12°\end{array}$