Chapter 12.4, Problem 34PPS

To determine

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

Answer to Problem 34PPS

In case 1

We got

  $\begin{array}{l}c=18.3\\ C={65}^0\\ B={71}^0\end{array}$

In case 2

We got

  $\begin{array}{l}c=9.2\\ C={27}^0\\ B={109}^0\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}A={44}^0\\ a=14\\ b=19\end{array}$

Given

  $\begin{array}{l}A={44}^0\\ a=14\\ b=19\end{array}$

Here $A={44}^0$ is Acute angle.

Because A is Acute and $14<19$ 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

  $\begin{array}{l}h=b\sin A\\ \Rightarrow h=19\sin {44}^0\\ \Rightarrow h=19\left(0.695\right)\\ \Rightarrow h=13.2\end{array}$

So, we got

  $h<a<b$

So, there will be Two solutions.

Case-1 Angle B is acute

Using law of sines

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin B}{b}\\ \Rightarrow \frac{\sin {44}^0}{14}=\frac{\sin B}{19}\\ \Rightarrow \sin B=\frac{19\sin {44}^0}{14}\\ \Rightarrow \sin B=\frac{19\left(0.695\right)}{14}\\ \Rightarrow \sin B=\frac{13.2}{14}\\ \Rightarrow \sin B=0.943\\ \Rightarrow B={\sin }^{-1}\left(0.943\right)\\ \Rightarrow B={71}^0\end{array}$

We know

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

  $\begin{array}{l}\Rightarrow A+B+C={180}^0\\ \Rightarrow {44}^0+{71}^0+C={180}^0\\ \Rightarrow {115}^0+C={180}^0\\ \Rightarrow C={180}^0-{115}^0\\ \Rightarrow C={65}^0\end{array}$

Using law of sines

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin C}{c}\\ \Rightarrow \frac{\sin {44}^0}{14}=\frac{\sin {65}^0}{c}\\ \Rightarrow c=\frac{14\sin {65}^0}{\sin {44}^0}\\ \Rightarrow c=\frac{14\left(0.906\right)}{0.695}\\ \Rightarrow c=\frac{12.684}{0.695}\\ \Rightarrow c=18.3\end{array}$

Therefore,

In case 1

We got

  $\begin{array}{l}c=18.3\\ C={65}^0\\ B={71}^0\end{array}$

Case-2 Angle B is obtuse

Using law of sines

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin B}{b}\\ \Rightarrow \frac{\sin {44}^0}{14}=\frac{\sin B}{19}\\ \Rightarrow \sin B=\frac{19\sin {44}^0}{14}\\ \Rightarrow \sin B=\frac{19\left(0.695\right)}{14}\\ \Rightarrow \sin B=\frac{13.2}{14}\\ \Rightarrow \sin B=0.943\\ \Rightarrow B={\sin }^{-1}\left(0.943\right)\end{array}$

  $\begin{array}{l}\Rightarrow B={180}^0-{71}^0\\ \Rightarrow B={109}^0\end{array}$

We know

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

  $\begin{array}{l}\Rightarrow A+B+C={180}^0\\ \Rightarrow {44}^0+{109}^0+C={180}^0\\ \Rightarrow {153}^0+C={180}^0\\ \Rightarrow C={180}^0-{153}^0\\ \Rightarrow C={27}^0\end{array}$

Using law of sines

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin C}{c}\\ \Rightarrow \frac{\sin {44}^0}{14}=\frac{\sin {27}^0}{c}\\ \Rightarrow c=\frac{14\sin {27}^0}{\sin {44}^0}\\ \Rightarrow c=\frac{14\left(0.454\right)}{0.695}\\ \Rightarrow c=\frac{6.356}{0.695}\\ \Rightarrow c=9.2\end{array}$

Therefore,

In case 2

We got

  $\begin{array}{l}c=9.2\\ C={27}^0\\ B={109}^0\end{array}$