Chapter 6, Problem 11RE

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 11RE

Thesolution of the triangle is $A={75}^0,B={39.48}^0,C={65.52}^0,a=51.2,b=33.7,c=48.24$ .

Explanation of Solution

Given info:

  $A={75}^0,a=51.2,b=33.7$

Formula used:

  

Law of sines

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

Calculation:

Given that $A={75}^0,a=51.2,b=33.7$ , We have

  $\begin{array}{l}A={75}^0,a=51.2,b=33.7\\ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\\ \frac{51.2}{\sin {75}^0}=\frac{33.7}{\sin B}=\frac{c}{\sin C}\\ \frac{33.7}{\sin B}=\frac{c}{\sin C}=53\\ \sin B=0.636\\ B={39.48}^0\text{ or }B={140.52}^0\text{ (not possible)}\\ C={180}^0-{75}^0-{39.48}^0={65.52}^0\\ \frac{c}{\sin {65.52}^0}=53\\ c=48.24\\ A={75}^0,B={39.48}^0,C={65.52}^0,a=51.2,b=33.7,c=48.24\end{array}$

Thesolution of the triangle is $A={75}^0,B={39.48}^0,C={65.52}^0,a=51.2,b=33.7,c=48.24$

Conclusion:

Thus,the solution of the triangle is $A={75}^0,B={39.48}^0,C={65.52}^0,a=51.2,b=33.7,c=48.24$ .