Chapter 6, Problem 5CT

To determine

To find: The solution for the given triangle.

Answer to Problem 5CT

The value of side of the triangle is $c=11.81$ and the angles of the triangle are $C=16.10°$ and $B=138.90°$ , the other pair of angles is $C=113.90°$ and $B=41.10°$ , the corresponding side is $c=38.94$ .

Explanation of Solution

Given:

The given sides of the triangle are $a=18$ and $b=28$ . The given angle are $A=25°$ .

Calculation:

The sketch for the triangle is shown in Figure 1

  

Figure 1

Consider the hypotenuse $h$ is,

  $h=b\sin A$

Then,

  $\begin{array}{c}h=28\sin \left(25°\right)\\ =11.83\end{array}$

Consider the law of sine as,

  $\begin{array}{l}\frac{a}{\sin A}=\frac{b}{\sin B}\\ \frac{\sin B}{b}=\frac{\sin A}{a}\\ \sin B=b\left(\frac{\sin A}{a}\right)\end{array}$

Then,

  $\begin{array}{l}\sin B=\left(28\right)\left(\frac{\sin 25°}{18}\right)\\ B=41.10°\end{array}$

Consider the case when $B=41.10°$ . Then by angle sum property,

  $\begin{array}{c}C=180°-25°-41.10°\\ =113.90°\end{array}$

Consider the law of sine as,

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

Then,

  $\begin{array}{l}c=\frac{\sin 113.90°}{\sin 25°}\\ c=38.94\end{array}$

Consider the case for the angle,

  $\begin{array}{c}B=180°-41.10°\\ =138.90°\end{array}$

Consider the case when $B=138.90°$ . Then by angle sum property,

  $\begin{array}{c}C=180°-25°-138.90°\\ =16.10°\end{array}$

Consider the law of sine as,

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

Then,

  $\begin{array}{c}c=\frac{\sin 16.10°}{\sin 25°}\\ =11.81\end{array}$