Chapter 13.4, Problem 31PPS

To determine

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

Answer to Problem 31PPS

One solution is

  $\begin{array}{l}B={32}^0\\ C={110}^0\\ c=32\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}A={38}^0\\ a=21\\ b=18\end{array}$

Given

  $\begin{array}{l}A={38}^0\\ a=21\\ b=18\end{array}$

Here $A={38}^0$ is Obtuse angle.

Because A is Obtuse and $21>18$ that is $a\ge b$ .

So, there is one solution.

Using law of sines

We get

  $\frac{\sin A}{a}=\frac{\sin B}{b}$

On substituting values

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin B}{b}\\ \Rightarrow \frac{\sin {38}^0}{21}=\frac{\sin B}{18}\\ \Rightarrow \sin B=\frac{18\sin {38}^0}{21}\\ \Rightarrow \sin B=\frac{18\left(0.616\right)}{21}\\ \Rightarrow \sin B=\frac{11.082}{21}\\ \Rightarrow \sin B=0.53\\ \Rightarrow B={\sin }^{-1}\left(0.53\right)\\ \Rightarrow B={32}^0\end{array}$

We know

Sum of Angles of Triangle is ${180}^0$

  $\begin{array}{l}A+B+C={180}^0\\ \Rightarrow {38}^0+{32}^0+C={180}^0\\ \Rightarrow {70}^0+C={180}^0\\ \Rightarrow C={180}^0-{70}^0\\ \Rightarrow C={110}^0\end{array}$

Using law of sines

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin C}{c}\\ \Rightarrow \frac{\sin {38}^0}{21}=\frac{\sin {110}^0}{c}\\ \Rightarrow c=\frac{21\sin {110}^0}{\sin {38}^0}\\ \Rightarrow c=\frac{21\left(0.94\right)}{0.616}\\ \Rightarrow c=\frac{19.74}{0.616}\\ \Rightarrow c=32\end{array}$

Therefore,

One solution is

  $\begin{array}{l}B={32}^0\\ C={110}^0\\ c=32\end{array}$