Chapter EP, Problem 13.4.17EP

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 13.4.17EP

The required values are for first set $B=30.5°$ , $124.5°$ , $c=243.7$ and the second set is $B=149.5°$ , $C=5.5°$ and $c=18.2$ this shows that there are two solution of the triangle ABC.

Explanation of Solution

Given:

The given data for triangle is $a=125$ , $b=150$ and $A=25°$ .

Calculation:

Since angle A is acute then,

  $\begin{array}{c}b\sin A=15\sin 25\\ =63.69\end{array}$

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

Then, by the acute angle property,

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin B}{b}\\ \sin B=\frac{150}{125}\left(\sin 25\right)\\ B={\sin }^{-1}0.507\\ B=30.5°\end{array}$

Consider by the angle sum property,

  $\begin{array}{l}A+B+C=180°\\ 55.55°+C=180°\\ C=124.5°\end{array}$

Then by the law of sine,

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin A}{c}\\ c=\frac{\sin 124.53}{\sin 25}125\\ c=243.7\end{array}$

For another case consider the angle B is acute and the obtuse angle whose since 0.507 is,

  $180°-30.5°=159.5°$

Then, by the angle sum property,

  $\begin{array}{l}A+B+C=180°\\ 25°+149.5°+C=180°\\ C=5.5°\end{array}$

Then by the law of sine,

  $\begin{array}{c}c=\frac{\sin C}{\sin A}a\\ =\frac{\sin 5.47}{\sin 25}\left(125\right)\\ =28.2\end{array}$

Thus, the required values are for first set $B=30.5°$ , $124.5°$ , $c=243.7$ and the second set is $B=149.5°$ , $C=5.5°$ and $c=18.2$ this shows that there are two solution of the triangle ABC.