Chapter 13.4, Problem 30PPS

To determine

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

Answer to Problem 30PPS

One solution is

  $\begin{array}{l}B={49}^0\\ C={56}^0\\ c=12.5\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}A={75}^0\\ a=14\\ b=11\end{array}$

Given

  $\begin{array}{l}A={75}^0\\ a=14\\ b=11\end{array}$

Here $A={75}^0$ is Obtuse angle.

Because A is Obtuse and $14>11$ 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

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin B}{b}\\ \Rightarrow \frac{\sin {75}^0}{14}=\frac{\sin B}{11}\\ \Rightarrow \sin B=\frac{11\sin {75}^0}{14}\\ \Rightarrow \sin B=\frac{11\left(0.966\right)}{14}\\ \Rightarrow \sin B=\frac{10.625}{14}\\ \Rightarrow \sin B=0.759\\ \Rightarrow B={\sin }^{-1}\left(0.759\right)\\ \Rightarrow B={49}^0\end{array}$

We know

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

  $\begin{array}{l}A+B+C={180}^0\\ \Rightarrow {75}^0+{49}^0+C={180}^0\\ \Rightarrow {124}^0+C={180}^0\\ \Rightarrow C={180}^0-{124}^0\\ \Rightarrow C={56}^0\end{array}$

Using law of sines

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin C}{c}\\ \Rightarrow \frac{\sin {75}^0}{14}=\frac{\sin {56}^0}{c}\\ \Rightarrow c=\frac{14\sin {56}^0}{\sin {75}^0}\\ \Rightarrow c=\frac{14\left(0.83\right)}{0.966}\\ \Rightarrow c=\frac{11.62}{0.966}\\ \Rightarrow c=12.5\end{array}$

Therefore,

One solution is

  $\begin{array}{l}B={49}^0\\ C={56}^0\\ c=12.5\end{array}$