Chapter 6, Problem 28CLT

To determine

To find: The solution for the given triangle.

Answer to Problem 28CLT

The value of angle $B=14.89°$ and angle $C=119.11°$ . The value of side $c=17$ .

Explanation of Solution

Given:

The given anglesis $A=46°$ . The sides are $a=14$ and $b=5$ .

The given diagram is shown in Figure 1

  

Figure 1

Calculation:

Consider the law of sine as,

  $\begin{array}{l}\frac{b}{\sin B}=\frac{a}{\sin A}\\ \sin B=b\frac{\sin A}{a}\end{array}$

Then,

  $\begin{array}{l}\sin B=5\frac{\sin 46°}{14}\\ B=14.89°\end{array}$

Consider the angle sum property of the triangle is,

  $C=180°-A-B$

Then,

  $\begin{array}{c}C=180°-46°-14.89°\\ =119.11°\end{array}$

Consider the law of sine as,

  $\begin{array}{l}\sin c=c\frac{\sin C}{A}\\ c=a\frac{\sin C}{\sin A}\end{array}$

Then,

  $\begin{array}{c}c=\left(14\right)\frac{\sin 119.11°}{\sin 46°}\\ =17\end{array}$