Chapter 12.4, Problem 8CYU

To determine

To find: Whether each triangle has no solution, one solution or two solutions.

Answer to Problem 8CYU

  $\begin{array}{l}c=13.7\\ B={39}^0\\ C={46}^0\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}A={95}^0\\ a=19\\ b=12\end{array}$

Given

  $\begin{array}{l}A={95}^0\\ a=19\\ b=12\end{array}$

Here $A={95}^0$ is Obtuse angle.

Because A is Obtuse and $19<12$ that is $a>b$

It means there is one solution.

Using law of sines

  $\begin{array}{l}\frac{\sin B}{b}=\frac{\sin A}{a}\\ \Rightarrow \frac{\sin B}{12}=\frac{\sin {95}^0}{19}\\ \Rightarrow \sin B=\frac{12\sin {95}^0}{19}\\ \Rightarrow \sin B=\frac{12\left(0.9962\right)}{19}\\ \Rightarrow \sin B=\frac{11.96}{19}\\ \Rightarrow \sin B=0.63\\ \Rightarrow B={\sin }^{-1}\left(0.63\right)\\ \Rightarrow B={39}^0\end{array}$

We know

Sum of angles of triangle $={180}^0$

Using that

We get

  $\begin{array}{l}A+B+C={180}^0\\ \Rightarrow 95+39+C={180}^0\\ \Rightarrow 134+C={180}^0\\ \Rightarrow C={180}^0-{134}^0\\ \Rightarrow C={46}^0\end{array}$

Again, using Laws of sines to find $c$

  $\begin{array}{l}\frac{\sin C}{c}=\frac{\sin A}{a}\\ \Rightarrow \frac{\sin {46}^0}{c}=\frac{\sin {95}^0}{19}\\ \Rightarrow c=\frac{19\sin {46}^0}{\sin {95}^0}\\ \Rightarrow c=\frac{19\left(0.72\right)}{\left(0.9962\right)}\\ \Rightarrow c=\frac{13.68}{0.9962}\\ \Rightarrow c=13.7\end{array}$

Therefore,

We got

  $\begin{array}{l}c=13.7\\ B={39}^0\\ C={46}^0\end{array}$