Chapter 13.4, Problem 36PPS

To determine

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

Answer to Problem 36PPS

  $\begin{array}{l}c=29.5\\ C={60}^0\\ B={90}^0\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}A={30}^0\\ a=17\\ b=34\end{array}$

Given

  $\begin{array}{l}A={30}^0\\ a=17\\ b=34\end{array}$

Here $A={30}^0$ is Acute angle.

Because A is Acute and $17<34$ that is $a<b$ .

So, we need to find $h$ and compare it to $a$

We know

  $h=b\sin A$

On substituting values

We get

  $\begin{array}{l}h=b\sin A\\ \Rightarrow h=34\sin {30}^0\\ \Rightarrow h=34\left(0.5\right)\\ \Rightarrow h=17\end{array}$

So, we got

  $a=h$

So, there will be One solution.

Using law of sines

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin B}{b}\\ \Rightarrow \frac{\sin {30}^0}{17}=\frac{\sin B}{34}\\ \Rightarrow \sin B=\frac{34\sin {30}^0}{17}\\ \Rightarrow \sin B=\frac{34\left(0.5\right)}{17}\\ \Rightarrow \sin B=\frac{17}{17}\\ \Rightarrow \sin B=1\\ \Rightarrow B={\sin }^{-1}\left(1\right)\\ \Rightarrow B={90}^0\end{array}$

We know

Sum of angles of a triangle is ${180}^0$

  $\begin{array}{l}\Rightarrow A+B+C={180}^0\\ \Rightarrow {30}^0+{90}^0+C={180}^0\\ \Rightarrow {120}^0+C={180}^0\\ \Rightarrow C={180}^0-{120}^0\\ \Rightarrow C={60}^0\end{array}$

Using law of sines

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin C}{c}\\ \Rightarrow \frac{\sin {30}^0}{17}=\frac{\sin {60}^0}{c}\\ \Rightarrow c=\frac{17\sin {60}^0}{\sin {30}^0}\\ \Rightarrow c=\frac{17\left(0.866\right)}{0.695}\\ \Rightarrow c=\frac{14.722}{0.5}\\ \Rightarrow c=29.5\end{array}$

Therefore,

We got

  $\begin{array}{l}c=29.5\\ C={60}^0\\ B={90}^0\end{array}$