Chapter EP, Problem 13.4.13EP

To determine

To Find: The number of solution that the triangle have and the value of the angle measure and the side measure.

Answer to Problem 13.4.13EP

The required values are, the first set is $a=26$ , $b=29$ and $A=58°$ , the second set is $B=108.9°$ , $C=13.1°$ and $c=6.9$ this shows that there are two solution of the triangle ABC.

Explanation of Solution

Given:

The given data for triangle is $a=26$ , $b=29$ and $A=58°$ .

Calculation:

Since angle A is acute then,

  $\begin{array}{c}b\sin A=26\sin 58\\ =22.049\end{array}$

The above $b>a>b\sin A$ there are two solutions for the given equations,

Then, by the acute angle property,

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin B}{b}\\ \sin B=\frac{29}{26}\left(\sin 25\right)\\ B={\sin }^{-1}0.946\\ B=71.1°\end{array}$

Consider by the angle sum property,

  $\begin{array}{l}A+B+C=180°\\ 71.1°+58°+C=180°\\ C=50.9°\end{array}$

Then by the law of sine,

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin A}{c}\\ c=\frac{\sin 50.93}{\sin 25}\left(26\right)\\ c=23.8\end{array}$

Therefore, the values are $B=71.1°$ , $C=50.9°$ and $c=23.8$ .

Since $\sin B=0.946$ .

For another case consider the angle B is acute and the obtuse angle whose since $0.946$ is,

  $180°-71.1°=108.9°$

Then, by the angle sum property,

  $\begin{array}{l}A+B+C=180°\\ 58°+108.9°+C=180°\\ C=13.1°\end{array}$

Then by the law of sine,

  $\begin{array}{c}c=\frac{\sin C}{\sin A}a\\ =\frac{\sin 13.07}{\sin 58}\left(26\right)\\ =6.9\end{array}$

Thus, the required values are, the first set is $a=26$ , $b=29$ and $A=58°$ , the second set is $B=108.9°$ , $C=13.1°$ and $c=6.9$ this shows that there are two solution of the triangle ABC.