Chapter 5.6, Problem 14E

To determine

To solve: Triangle $A={65}^{\circ },B={50}^{\circ },c=12$ . Round to the nearest tenth

Answer to Problem 14E

  $\angle C={65}^{\circ },a=12,b=10.1$

Explanation of Solution

Given information: $A={65}^{\circ },B={50}^{\circ },c=12$

Calculation:

  $\angle C=180-\left(25+35\right)={120}^{\circ }$

So this is an isosceles triangle so c and an are equal. We only need to find b .

Law if sines:

  $\begin{array}{l}\frac{b}{\sin {50}^{\circ }}=\frac{12}{\sin {65}^{\circ }}\\ b=\frac{12\times \sin {50}^{\circ }}{\sin {65}^{\circ }}=10.1\end{array}$

Thus,

  $\angle C={65}^{\circ },a=12,b=10.1$