Chapter 12.4, Problem 10CYU

To determine

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

Answer to Problem 10CYU

In case 1

We got

  $\begin{array}{l}c=14.1\\ C={80}^0\\ B={66}^0\end{array}$

In case 2

We got

  $\begin{array}{l}c=7.6\\ C={32}^0\\ B={114}^0\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}A={34}^0\\ a=8\\ b=13\end{array}$

Given

  $\begin{array}{l}A={34}^0\\ a=8\\ b=13\end{array}$

Here $A={34}^0$ is Acute angle.

Because A is Acute and $8<13$ 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

  $\begin{array}{l}h=b\sin A\\ \Rightarrow h=13\sin \left({34}^0\right)\\ \Rightarrow h=13\left(0.56\right)\\ \Rightarrow h=7.27\end{array}$

Therefore,

We got $h<a<b$

It means there will be two Solutions.

Case 1

Angle B is acute.

Use law of sines

We get

  $\begin{array}{l}\frac{\sin B}{b}=\frac{\sin A}{a}\\ \Rightarrow \frac{\sin B}{13}=\frac{\sin \left({34}^0\right)}{8}\\ \Rightarrow \frac{\sin B}{13}=\frac{\left(0.56\right)}{8}\\ \Rightarrow \sin B=\frac{13\left(0.56\right)}{8}\\ \Rightarrow \sin B=\frac{7.28}{8}\\ \Rightarrow \sin B=0.91\\ \Rightarrow B={\sin }^{-1}\left(0.91\right)\\ \Rightarrow B={66}^0\end{array}$

We know

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

  $\begin{array}{l}A+B+C={180}^0\\ \Rightarrow {34}^0+{66}^0+C={180}^0\\ \Rightarrow {100}^0+C={180}^0\\ \Rightarrow C={180}^0-{100}^0\\ \Rightarrow C={80}^0\end{array}$

Using law of Sines

We get

  $\begin{array}{l}\frac{\sin C}{c}=\frac{\sin A}{a}\\ \Rightarrow \frac{\sin {80}^0}{c}=\frac{\sin {34}^0}{8}\\ \Rightarrow c=\frac{8\sin {80}^0}{\sin {34}^0}\\ \Rightarrow c=\frac{8\left(0.985\right)}{\left(0.56\right)}\\ \Rightarrow c=\frac{7.88}{\left(0.56\right)}\\ \Rightarrow c=14.1\end{array}$

Therefore,

In case 1

We got

  $\begin{array}{l}c=14.1\\ C={80}^0\\ B={66}^0\end{array}$

Case 2

Angle B is obtuse.

Use law of sines

We get

  $\begin{array}{l}\frac{\sin B}{b}=\frac{\sin A}{a}\\ \Rightarrow \frac{\sin B}{13}=\frac{\sin \left({34}^0\right)}{8}\\ \Rightarrow \frac{\sin B}{13}=\frac{\left(0.56\right)}{8}\\ \Rightarrow \sin B=\frac{13\left(0.56\right)}{8}\\ \Rightarrow \sin B=\frac{7.28}{8}\\ \Rightarrow \sin B=0.91\\ \Rightarrow B={\sin }^{-1}\left(0.91\right)\end{array}$

  $\begin{array}{l}\Rightarrow B={180}^0-{66}^0\\ \Rightarrow B={114}^0\end{array}$

We know

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

  $\begin{array}{l}A+B+C={180}^0\\ \Rightarrow {34}^0+{114}^0+C={180}^0\\ \Rightarrow {148}^0+C={180}^0\\ \Rightarrow C={180}^0-{148}^0\\ \Rightarrow C={32}^0\end{array}$

Using law of Sines

We get

  $\begin{array}{l}\frac{\sin C}{c}=\frac{\sin A}{a}\\ \Rightarrow \frac{\sin {32}^0}{c}=\frac{\sin {34}^0}{8}\\ \Rightarrow c=\frac{8\sin {32}^0}{\sin {34}^0}\\ \Rightarrow c=\frac{8\left(0.53\right)}{\left(0.56\right)}\\ \Rightarrow c=\frac{4.24}{\left(0.56\right)}\\ \Rightarrow c=7.6\end{array}$

Therefore,

In case 2

We got

  $\begin{array}{l}c=7.6\\ C={32}^0\\ B={114}^0\end{array}$