Chapter 8, Problem 28RE

To determine

To find: The remaining angle(s) and side(s) of the triangle, if it(they) exist.

Answer to Problem 28RE

The two possible triangle measurements are $c=4.54$ , $B=30.87°$ , $C=129.13°$ and $c=1.1$ , $B=149.13°$ , $C=10.87°$ .

Explanation of Solution

Given:

The sides $a=2$ , $b=3$ and the angle $A=20°$ .

Calculation:

The Sine Law for a triangle $\Delta ABC$ is given by $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$ , where capital letters represent the angles and the small letters represent the length of the respective sides of the triangle.

Given the sides a and b , and the angle A , angle B can be determined as follows:

  $\begin{array}{c}\frac{\sin A}{a}=\frac{\sin B}{b}\\ \frac{\sin 20°}{2}=\frac{\sin B}{3}\\ \sin B=\frac{3\sin 20°}{2}\\ B=30.87°,149.13°\end{array}$

Since there are 2 possible values for the angle B , two possible triangles exist.

Case I: $B=30.87°$

With the two angles known, the third angle C is determined as shown below:

  $\begin{array}{c}A+B+C=180°\\ 20°+30.87°+C=180°\\ C=180°-50.87°\\ =129.13°\end{array}$

The third side c is determined using the Sine Law as shown below:

  $\begin{array}{c}\frac{\sin A}{a}=\frac{\sin C}{c}\\ \frac{\sin 20°}{2}=\frac{\sin 129.13°}{c}\\ c=\frac{2\sin 129.13°}{\sin 20°}\\ =4.54\end{array}$

Case II: $B=149.13°$

With the two angles known, the third angle C is determined as shown below:

  $\begin{array}{c}A+B+C=180°\\ 20°+149.13°+C=180°\\ C=180°-169.13°\\ =10.87°\end{array}$

The third side c is determined using the Sine Law as shown below:

  $\begin{array}{c}\frac{\sin A}{a}=\frac{\sin C}{c}\\ \frac{\sin 20°}{2}=\frac{\sin 10.87°}{c}\\ c=\frac{2\sin 10.87°}{\sin 20°}\\ =1.1\end{array}$

So, the two possible triangle measurements are $c=4.54$ , $B=30.87°$ , $C=129.13°$ and $c=1.1$ , $B=149.13°$ , $C=10.87°$ .