Chapter 13.4, Problem 11CYU

To determine

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

Answer to Problem 11CYU

  $\begin{array}{l}c=5.2\\ C={60}^0\\ B={90}^0\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}A={30}^0\\ a=3\\ b=6\end{array}$

Given

  $\begin{array}{l}A={30}^0\\ a=3\\ b=6\end{array}$

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

Because A is Acute and $3<6$ that is $a<b$

Now we need to find $h$ and compare it to $a$ .

We know

  $h=b\sin A$

On substituting values

We get

  $\begin{array}{l}h=b\sin A\\ \Rightarrow h=6\sin \left({30}^0\right)\\ \Rightarrow h=6\left(0.5\right)\\ \Rightarrow h=3\end{array}$

Therefore,

We got $a=h$

It means there will be One Solution.

Use law of sines

We get

  $\begin{array}{l}\frac{\sin B}{b}=\frac{\sin A}{a}\\ \Rightarrow \frac{\sin B}{6}=\frac{\sin \left({30}^0\right)}{3}\\ \Rightarrow \frac{\sin B}{6}=\frac{\left(0.5\right)}{3}\\ \Rightarrow \sin B=\frac{6\left(0.5\right)}{3}\\ \Rightarrow \sin B=\frac{3}{3}\\ \Rightarrow \sin B=1\\ \Rightarrow B={\sin }^{-1}\left(1\right)\\ \Rightarrow B={90}^0\end{array}$

We know

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

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

Using law of Sines

We get

  $\begin{array}{l}\frac{\sin C}{c}=\frac{\sin A}{a}\\ \Rightarrow \frac{\sin {60}^0}{c}=\frac{\sin {30}^0}{3}\\ \Rightarrow c=\frac{3\sin {60}^0}{\sin {30}^0}\\ \Rightarrow c=\frac{3\left(0.866\right)}{0.5}\\ \Rightarrow c=\frac{2.6}{0.5}\\ \Rightarrow c=5.2\end{array}$

Therefore,

We got

  $\begin{array}{l}c=5.2\\ C={60}^0\\ B={90}^0\end{array}$