Chapter 6.2, Problem 12E

To determine

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

Answer to Problem 12E

The angles of the triangle area $A=42.06°$ , $B=12.91°$ and $C=125.03°$ .

Explanation of Solution

Given:

The sides of the triangle are,

  $\begin{array}{l}a=9\\ b=3\\ c=11\end{array}$

Calculation:

Consider the alternative form of Law of cosine is,

  $\cos C=\frac{a^2+b^2-c^2}{2ab}$

Then,

  $\begin{array}{l}\cos C=\frac{{\left(9\right)}^2+{\left(3\right)}^2-{\left(11\right)}^2}{2\left(9\right)\left(3\right)}\\ \cos C=-0.741\\ C=125.03°\end{array}$

Consider the law of sine as,

  $\sin A=a\left(\frac{\sin C}{c}\right)$

Then,

  $\begin{array}{l}\sin A=\left(9\right)\left(\frac{\sin 25.03°}{11}\right)\\ A=42.06°\end{array}$

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

  $B=180°-A-C$

Then,

  $\begin{array}{c}B=180°-125.03°-42.06°\\ =12.91°\end{array}$