Chapter 8.2, Problem 11AYU

In Problems 9-16, solve each triangle.

To determine

To find: The remaining angles and sides of the given triangle.

Answer to Problem 11AYU

Solution:

$\text{a} \approx 3.25, \text{c} = 4.23$ and $\text{B} = {45}^{\circ }$

Explanation of Solution

Given:

Formula used:

$\frac{\text{sin}\text{A}}{\text{a}} = \frac{\text{sin}\text{B}}{\text{b}} = \frac{\text{sin}\text{C}}{\text{c}}$

$\angle \text{A} + \angle \text{B} + \angle \text{C} = {180}^{\circ }$

Calculation:

$\begin{array}{l}\angle \text{A} + \angle \text{B} + \angle \text{C} = {180}^{\circ }\\ {50}^{\circ } + {85}^{\circ } + \angle \text{C} = {180}^{\circ }\\ \angle \text{C} = {180}^{\circ } - {85}^{\circ } - {50}^{\circ }\\ \angle \text{C} = {45}^{\circ }\end{array}$

The Law of Sines can now be used to find the two distances $\text{a}$ and $\text{b}$ that are needed.

$\frac{\text{sin}\text{A}}{\text{a}} = \frac{\text{sin}\text{B}}{\text{b}} = \frac{\text{sin}\text{C}}{\text{c}}$

$\frac{\text{sin}A}{a} = \frac{\text{sin}B}{b}$

$\frac{\text{sin}50°}{a} = \frac{\text{sin}45°}{3}$

$a = 3 * \frac{\text{sin}50°}{\text{sin}45°}$

$a = 3.25$

$\frac{\text{sin}B}{b} = \frac{\text{sin}C}{c}$

$\frac{\text{sin}45°}{3} = \frac{\text{sin}85°}{c}$

$c = 3 * \frac{\text{sin}85°}{\text{sin}45°}$

$c = 4.23$