Chapter 12.4, Problem 33PPS

To determine

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

Answer to Problem 33PPS

In case 1

We got

  $\begin{array}{l}c=7.4\\ C={85}^0\\ B={53}^0\end{array}$

In case 2

We got

  $\begin{array}{l}c=1.4\\ C={11}^0\\ B={127}^0\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}A={42}^0\\ a=5\\ b=6\end{array}$

Given

  $\begin{array}{l}A={42}^0\\ a=5\\ b=6\end{array}$

Here $A={42}^0$ is Obtuse angle.

Because A is Obtuse and $5<6$ 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

  $\begin{array}{l}h=b\sin A\\ \Rightarrow h=6\sin {42}^0\\ \Rightarrow h=6\left(0.67\right)\\ \Rightarrow h=4\end{array}$

So, we got

  $h<a<b$

So, there will be Two solutions.

Case-1 Angle B is acute

Using law of sines

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin B}{b}\\ \Rightarrow \frac{\sin {42}^0}{5}=\frac{\sin B}{6}\\ \Rightarrow \sin B=\frac{6\sin {42}^0}{5}\\ \Rightarrow \sin B=\frac{6\left(0.67\right)}{5}\\ \Rightarrow \sin B=\frac{4}{5}\\ \Rightarrow \sin B=0.8\\ \Rightarrow B={\sin }^{-1}\left(0.8\right)\\ \Rightarrow B={53}^0\end{array}$

We know

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

  $\begin{array}{l}\Rightarrow A+B+C={180}^0\\ \Rightarrow {42}^0+{53}^0+C={180}^0\\ \Rightarrow {95}^0+C={180}^0\\ \Rightarrow C={180}^0-{95}^0\\ \Rightarrow C={85}^0\end{array}$

Using law of sines

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin C}{c}\\ \Rightarrow \frac{\sin {42}^0}{5}=\frac{\sin {85}^0}{c}\\ \Rightarrow c=\frac{5\sin {85}^0}{\sin {42}^0}\\ \Rightarrow c=\frac{5\left(0.996\right)}{0.669}\\ \Rightarrow c=\frac{4.98}{0.669}\\ \Rightarrow c=7.4\end{array}$

Therefore,

In case 1

We got

  $\begin{array}{l}c=7.4\\ C={85}^0\\ B={53}^0\end{array}$

Case-2 Angle B is obtuse

Using law of sines

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin B}{b}\\ \Rightarrow \frac{\sin {42}^0}{5}=\frac{\sin B}{6}\\ \Rightarrow \sin B=\frac{6\sin {42}^0}{5}\\ \Rightarrow \sin B=\frac{6\left(0.67\right)}{5}\\ \Rightarrow \sin B=\frac{4}{5}\\ \Rightarrow \sin B=0.8\\ \Rightarrow B={\sin }^{-1}\left(0.8\right)\end{array}$

  $\begin{array}{l}\Rightarrow B={180}^0-{53}^0\\ \Rightarrow B={127}^0\end{array}$

We know

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

  $\begin{array}{l}\Rightarrow A+B+C={180}^0\\ \Rightarrow {42}^0+{127}^0+C={180}^0\\ \Rightarrow {169}^0+C={180}^0\\ \Rightarrow C={180}^0-{169}^0\\ \Rightarrow C={11}^0\end{array}$

Using law of sines

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin C}{c}\\ \Rightarrow \frac{\sin {42}^0}{5}=\frac{\sin {11}^0}{c}\\ \Rightarrow c=\frac{5\sin {11}^0}{\sin {42}^0}\\ \Rightarrow c=\frac{5\left(0.19\right)}{0.669}\\ \Rightarrow c=\frac{0.954}{0.669}\\ \Rightarrow c=1.4\end{array}$

Therefore,

In case 2

We got

  $\begin{array}{l}c=1.4\\ C={11}^0\\ B={127}^0\end{array}$