Chapter 6, Problem 12RE

To determine

Whether the Law of Sines is needed to solve the triangle. Then solve (if possible) the triangle. If two solutions exist, find both. Round your answers to two decimal places.

Answer to Problem 12RE

The solutions of the triangle are

  $\begin{array}{l}A={40.92}^0,B={25}^0,C={114.08}^0,a=6.2,b=4,c=8.64\text{ and}\\ A={40.92}^0,B={25}^0,C={15.92}^0,a=6.2,b=4,c=2.59\end{array}$

Explanation of Solution

Given info:

  $B={25}^0,a=6.2,b=4$

Formula used:

  

Law of sines

  $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

Calculation:

Given that $B={25}^0,a=6.2,b=4$ , We have

  $\begin{array}{l}B={25}^0,a=6.2,b=4\\ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\\ \frac{6.2}{\sin A}=\frac{4}{\sin {25}^0}=\frac{c}{\sin C}\\ \frac{6.2}{\sin A}=\frac{c}{\sin C}=9.46\\ \sin A=0.655\\ A={40.92}^0\text{ or }A={139.08}^0\\ \underset{\_}{\text{if }A={40.92}^0}\\ C={180}^0-{25}^0-{40.92}^0={114.08}^0\\ \frac{c}{\sin {114.08}^0}=9.46\\ c=8.64\\ A={40.92}^0,B={25}^0,C={114.08}^0,a=6.2,b=4,c=8.64\\ \underset{\_}{\text{if }A={139.08}^0}\\ C={180}^0-{25}^0-{139.08}^0={15.92}^0\\ \frac{c}{\sin {15.92}^0}=9.46\\ c=2.59\\ A={40.92}^0,B={25}^0,C={15.92}^0,a=6.2,b=4,c=2.59\end{array}$

The solutions of the triangle are

  $\begin{array}{l}A={40.92}^0,B={25}^0,C={114.08}^0,a=6.2,b=4,c=8.64\text{ and}\\ A={40.92}^0,B={25}^0,C={15.92}^0,a=6.2,b=4,c=2.59\end{array}$

Conclusion:

Thus,the solutions of the triangle are

  $\begin{array}{l}A={40.92}^0,B={25}^0,C={114.08}^0,a=6.2,b=4,c=8.64\text{ and}\\ A={40.92}^0,B={25}^0,C={15.92}^0,a=6.2,b=4,c=2.59\end{array}$