Chapter 12.6, Problem 51SR

To determine

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

Answer to Problem 51SR

One solution is

  $\begin{array}{l}B={35}^0\\ C={99}^0\\ c=13.7\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}A={46}^0\\ a=10\\ b=8\end{array}$

Given

  $\begin{array}{l}A={46}^0\\ a=10\\ b=8\end{array}$

Here Angle A is acute and

  $10>8$ it means $a>b$

So, there will be one solution.

Using law of sines

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin B}{b}\\ \Rightarrow \frac{\sin {46}^0}{10}=\frac{\sin B}{8}\\ \Rightarrow \sin B=\frac{8\sin {46}^0}{10}\\ \Rightarrow \sin B=\frac{8\left(0.72\right)}{10}\\ \Rightarrow \sin B=\frac{5.76}{10}\\ \Rightarrow \sin B=0.576\\ \Rightarrow B={\sin }^{-1}\left(0.576\right)\\ \Rightarrow B={35}^0\end{array}$

We know

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

Using that

We get

  $\begin{array}{l}A+B+C={180}^0\\ \Rightarrow {46}^0+{35}^0+C={180}^0\\ \Rightarrow {81}^0+C={180}^0\\ \Rightarrow C={180}^0-{81}^0\\ \Rightarrow C={99}^0\end{array}$

Again, Using law of sines

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin C}{c}\\ \Rightarrow \frac{\sin {46}^0}{10}=\frac{\sin {99}^0}{c}\\ \Rightarrow c=\frac{10\sin {99}^0}{\sin {46}^0}\\ \Rightarrow c=\frac{10\left(0.9877\right)}{0.72}\\ \Rightarrow c=\frac{9.877}{0.72}\\ \Rightarrow c=13.7\end{array}$

Therefore,

One solution is

  $\begin{array}{l}B={35}^0\\ C={99}^0\\ c=13.7\end{array}$