Chapter 12.4, Problem 3CGP

To determine

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

Answer to Problem 3CGP

In case 1

We got

  $\begin{array}{l}c=19.5\\ C={72}^0\\ B={61}^0\end{array}$

In case 2

We got

  $\begin{array}{l}c=4.96\\ C={14}^0\\ B={119}^0\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}A={47}^0\\ a=15\\ b=18\end{array}$

Given

  $\begin{array}{l}A={47}^0\\ a=15\\ b=18\end{array}$

Here $A={47}^0$ is Acute angle.

Because A is Acute and $15<18$ that is $a<b$

We need find $h$ and compare it to $a$ .

We know

  $\begin{array}{l}h=b\sin A\\ \Rightarrow h=18\sin \left({47}^0\right)\\ \Rightarrow h=18\left(0.7314\right)\\ \Rightarrow h=13.16\end{array}$

Therefore

We got

  $h<a<b$

It means there are two solutions.

Case 1

Angle A is acute

Using law of sines

  $\begin{array}{l}\\ \frac{\sin B}{b}=\frac{\sin A}{a}\\ \Rightarrow \frac{\sin B}{18}=\frac{\sin {47}^0}{15}\\ \Rightarrow \sin B=\frac{18\sin {47}^0}{15}\\ \Rightarrow \sin B=\frac{18\left(0.731\right)}{15}\\ \Rightarrow \sin B=\frac{13.1644}{15}\\ \Rightarrow \sin B=0.8776\\ \Rightarrow B={\sin }^{-1}\left(0.8776\right)\\ \Rightarrow B={61}^0\end{array}$

We know

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

Using that

We get

  $\begin{array}{l}A+B+C={180}^0\\ \Rightarrow 47+61+C={180}^0\\ \Rightarrow 108+C={180}^0\\ \Rightarrow C={180}^0-{108}^0\\ \Rightarrow C={72}^0\end{array}$

Again, using Laws of sines to find $c$

  $\begin{array}{l}\frac{\sin C}{c}=\frac{\sin A}{a}\\ \Rightarrow \frac{\sin {72}^0}{c}=\frac{\sin {47}^0}{15}\\ \Rightarrow c=\frac{15\sin {72}^0}{\sin {47}^0}\\ \Rightarrow c=\frac{15\left(0.951\right)}{\left(0.7314\right)}\\ \Rightarrow c=\frac{14.266}{\left(0.7314\right)}\\ \Rightarrow c=19.5\end{array}$

In case 1

We got

  $\begin{array}{l}c=19.5\\ C={72}^0\\ B={61}^0\end{array}$

Case 2

Angle B is obtuse.

The sine function also has a positive value in Quadrant-II.

So,

  $\begin{array}{l}\\ \frac{\sin B}{b}=\frac{\sin A}{a}\\ \Rightarrow \frac{\sin B}{18}=\frac{\sin {47}^0}{15}\\ \Rightarrow \sin B=\frac{18\sin {47}^0}{15}\\ \Rightarrow \sin B=\frac{18\left(0.731\right)}{15}\\ \Rightarrow \sin B=\frac{13.1644}{15}\\ \Rightarrow \sin B=0.8776\\ \Rightarrow B={\sin }^{-1}\left(0.8776\right)\end{array}$

  $\begin{array}{l}\Rightarrow B={180}^0-{61}^0\\ \Rightarrow B={119}^0\end{array}$

We know

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

Using that

We get

  $\begin{array}{l}A+B+C={180}^0\\ \Rightarrow 47+119+C={180}^0\\ \Rightarrow 166+C={180}^0\\ \Rightarrow C={180}^0-{166}^0\\ \Rightarrow C={14}^0\end{array}$

Again, using Laws of sines to find $c$

  $\begin{array}{l}\frac{\sin C}{c}=\frac{\sin A}{a}\\ \Rightarrow \frac{\sin {14}^0}{c}=\frac{\sin {47}^0}{15}\\ \Rightarrow c=\frac{15\sin {14}^0}{\sin {47}^0}\\ \Rightarrow c=\frac{15\left(0.242\right)}{\left(0.7314\right)}\\ \Rightarrow c=\frac{3.629}{\left(0.7314\right)}\\ \Rightarrow c=4.96\end{array}$

Therefore,

In case 2

We got

  $\begin{array}{l}c=4.96\\ C={14}^0\\ B={119}^0\end{array}$