Chapter 5.6, Problem 6CFU

To determine

To solve: Triangle $a=8.6,A={27.3}^{\circ },B={55.9}^{\circ }$ . Round to the nearest tenth

Answer to Problem 6CFU

  $m\angle C={96.8}^{\circ },b=15.5,c=18.6$

Explanation of Solution

Given information: $a=8.6,A={27.3}^{\circ },B={55.9}^{\circ }$

Calculation:

We are given $\angle A={27.3}^{\circ },\angle B={55.9}^{\circ }$ and a = 8.6

First find $m\angle C$

  $m\angle C=180-\left(27.3+55.9\right)={96.8}^{\circ }$

Use the law of sines:

  $\begin{array}{l}\frac{8.6}{\sin {27.3}^{\circ }}=\frac{b}{\sin {55.9}^{\circ }}=\frac{c}{\sin {96.8}^{\circ }}\\ b=\frac{8.6\times \sin {55.9}^{\circ }}{\sin {27.3}^{\circ }}\text{ and c}=\frac{8.6\times \sin {96.8}^{\circ }}{\sin {27.3}^{\circ }}\\ b=15.5,c=18.6\end{array}$

Thus,

  $m\angle C={96.8}^{\circ },b=15.5,c=18.6$