Chapter 5.6, Problem 24E

To determine

To find: The area of triangle $a=19.2,A={53.8}^{\circ },C={65.4}^{\circ }$ . Round to the nearest tenth

Answer to Problem 24E

  $\text{Area}=181.3u^2$

Explanation of Solution

Given information: $a=19.2,A={53.8}^{\circ },C={65.4}^{\circ }$

Calculation:

We have two formulas. It depends on which information we are given as to which one we use.

If we are given two sides and the included angle we use,

  $k=\frac{1}{2}bc\sin A$

If we are given (or can figure out) three angles and a side we use,

  $k=\frac{1}{2}{b}^2\frac{\sin A\times \sin C}{\sin B}$

In this problem we are given one side and two angles, so we use the second equation,

First we need to find the angle B .

  $\angle B=180-\left(53.8+65.4\right)={60.8}^{\circ }$

  $\begin{array}{l}k=\frac{1}{2}{a}^2\frac{\sin B\times \sin C}{\sin A}\\ k=\frac{1}{2}{19.2}^2\frac{\sin {60.8}^{\circ }\times \sin {65.4}^{\circ }}{\sin {53.8}^{\circ }}\\ k=0.5\times 368.64\times 0.9836\\ k=181.3\end{array}$

Thus,

  $\text{Area}=181.3u^2$