Chapter 13, Problem 30SGR

To determine

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

Answer to Problem 30SGR

One solution is

  $\begin{array}{l}B={41}^0\\ A={21}^0\\ b=7.4\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}C={118}^0\\ c=10\\ a=4\end{array}$

Given

  $\begin{array}{l}C={118}^0\\ c=10\\ a=4\end{array}$

Here $C={118}^0$ is Obtuse angle.

Because A is Obtuse and $10>4$ that is $c>a$ .

So, there is one solution.

Using law of sines

We get

  $\frac{\sin A}{a}=\frac{\sin C}{c}$

On substituting values

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin C}{c}\\ \Rightarrow \frac{\sin A}{4}=\frac{\sin {118}^0}{10}\\ \Rightarrow \sin A=\frac{4\sin {118}^0}{10}\\ \Rightarrow \sin A=\frac{4\left(0.883\right)}{10}\\ \Rightarrow \sin A=\frac{3.532}{10}\\ \Rightarrow \sin A=0.3532\\ \Rightarrow A={\sin }^{-1}\left(0.3532\right)\\ \Rightarrow A={21}^0\end{array}$

We know

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

  $\begin{array}{l}A+B+C={180}^0\\ \Rightarrow {21}^0+B+{118}^0={180}^0\\ \Rightarrow {139}^0+B={180}^0\\ \Rightarrow C={180}^0-{139}^0\\ \Rightarrow C={41}^0\end{array}$

Using law of sines

We get

  $\begin{array}{l}\frac{\sin B}{b}=\frac{\sin C}{c}\\ \Rightarrow \frac{\sin {41}^0}{b}=\frac{\sin {118}^0}{10}\\ \Rightarrow b=\frac{10\sin {41}^0}{\sin {118}^0}\\ \Rightarrow b=\frac{10\left(0.656\right)}{0.883}\\ \Rightarrow b=\frac{6.56}{0.883}\\ \Rightarrow b=7.4\end{array}$

Therefore,

One solution is

  $\begin{array}{l}B={41}^0\\ A={21}^0\\ b=7.4\end{array}$