Chapter 5.6, Problem 12E

To determine

To solve: Triangle $B={100}^{\circ },C={50}^{\circ },c=30$ . Round to the nearest tenth

Answer to Problem 12E

  $b=38.6,a=19.6,\angle A={30}^{\circ }$

Explanation of Solution

Given information: $B={100}^{\circ },C={50}^{\circ },c=30$

Calculation:

  $A=180-\left(100+50\right)={30}^{\circ }$

  $\begin{array}{l}\frac{a}{\sin {30}^{\circ }}=\frac{b}{\sin {100}^{\circ }}=\frac{30}{\sin {50}^{\circ }}\\ b=\frac{30\times \sin {100}^{\circ }}{\sin {50}^{\circ }}=38.6\\ a=\frac{30\times \sin {30}^{\circ }}{\sin {50}^{\circ }}=19.6\end{array}$

Thus,

  $b=38.6,a=19.6,\angle A={30}^{\circ }$