Chapter 13.4, Problem 29PPS

To determine

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

Answer to Problem 29PPS

One solution is

  $\begin{array}{l}B={25}^0\\ C={55}^0\\ c=5.8\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}A={100}^0\\ a=7\\ b=3\end{array}$

Given

  $\begin{array}{l}A={100}^0\\ a=7\\ b=3\end{array}$

Here $A={100}^0$ is Obtuse angle.

Because A is Obtuse and $7>3$ that is $a>b$ .

So, there is one solution.

Using law of sines

We get

  $\frac{\sin A}{a}=\frac{\sin B}{b}$

On substituting values

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin B}{b}\\ \Rightarrow \frac{\sin {100}^0}{7}=\frac{\sin B}{3}\\ \Rightarrow \sin B=\frac{3\sin {100}^0}{7}\\ \Rightarrow \sin B=\frac{3\left(0.9848\right)}{7}\\ \Rightarrow \sin B=\frac{2.9544}{7}\\ \Rightarrow \sin B=0.422\\ \Rightarrow B={\sin }^{-1}\left(0.422\right)\\ \Rightarrow B={25}^0\end{array}$

We know

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

  $\begin{array}{l}A+B+C={180}^0\\ \Rightarrow {100}^0+{25}^0+C={180}^0\\ \Rightarrow {125}^0+C={180}^0\\ \Rightarrow C={180}^0-{125}^0\\ \Rightarrow C={55}^0\end{array}$

Using law of sines

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin C}{c}\\ \Rightarrow \frac{\sin {100}^0}{7}=\frac{\sin {55}^0}{c}\\ \Rightarrow c=\frac{7\sin {55}^0}{\sin {100}^0}\\ \Rightarrow c=\frac{7\left(0.82\right)}{0.9848}\\ \Rightarrow c=\frac{5.74}{0.9848}\\ \Rightarrow c=5.8\end{array}$

Therefore,

One solution is

  $\begin{array}{l}B={25}^0\\ C={55}^0\\ c=5.8\end{array}$