Chapter 5.8, Problem 15E

To determine

To solve: the triangle, round to the nearest tenth.

Answer to Problem 15E

  $A=51.8°;B=70.9°\text{ and }C=57.3°$

Explanation of Solution

Given information:

  $a=11.4,b=13.7\text{ and }c=12.2$

Calculation:

The triangle will be as follows:

  

Evaluate the angle $B$ as follows:

  $\begin{array}{c}{b}^2=a^2+c^2-2\left(a\right)\left(c\right)\cos B\\ {13.7}^2={11.4}^2+{12.2}^2-2\left(11.4\right)\left(12.2\right)\cos B\\ \cos B=\frac{{13.7}^2-\left({11.4}^2+{12.2}^2\right)}{\left[-2\left(11.4\right)\left(12.2\right)\right]}\\ B={\cos }^{-1}\left[\frac{{13.7}^2-\left({11.4}^2+{12.2}^2\right)}{\left[-2\left(11.4\right)\left(12.2\right)\right]}\right]\\ =70.9°\end{array}$

Now evaluate angle $A$ as follows:

  $\begin{array}{c}\frac{a}{\sin A}=\frac{b}{\sin B}\\ \frac{11.4}{\sin A}=\frac{13.7}{\sin 70.9°}\\ \sin A=\frac{11.4}{13.7}\left(\sin 70.9°\right)\\ A={\sin }^{-1}\left(\frac{11.4\left(\sin 70.9°\right)}{13.7}\right)\\ =51.8°\end{array}$

Use angle sum property of triangle to evaluate angle $A$ as shown below.

  $\begin{array}{c}A+B+C=180°\\ 70.9°+51.8°+C=180°\\ C=180°-70.9°-51.8°\\ =57.3°\end{array}$

Thus, $A,B\text{ and }C$ the value of is $51.8°,70.9°\text{ and }57.3°$ respectively.