Chapter 6.1, Problem 11E

To determine

The solution of triangle using law of sines given that $A={83}^{0}20\text{'},C={54.6}^0,c=18.1$ .

Answer to Problem 11E

Thesolution is $A={83}^{0}20\text{'},B={42.07}^0,C={54.6}^0,a=22.05,b=14.88,c=18.1$

Explanation of Solution

Formula used:

Given info:

  $A={83}^{0}20\text{'},C={54.6}^0,c=18.1$

Formula used:

  

Law of sines

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

Calculation:

Given that $A={83}^{0}20\text{'},C={54.6}^0,c=18.1$ ,

We have

  $\begin{array}{l}\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\\ B={180}^0-({83}^{0}20\text{'}+{54.6}^0)={42.07}^0\\ \frac{a}{\sin {83}^{0}20\text{'}}=\frac{b}{\sin {42.07}^0}=\frac{18.1}{\sin {54.6}^0}\\ \frac{a}{\text{0}\text{.9932}}=\frac{b}{0.67}=\frac{18.1}{0.8151}\\ a=22.05\\ b=14.88\\ A={83}^{0}20\text{'},B={42.07}^0,C={54.6}^0,a=22.05,b=14.88,c=18.1\end{array}$

Conclusion:

Thus,the solution is $A={83}^{0}20\text{'},B={42.07}^0,C={54.6}^0,a=22.05,b=14.88,c=18.1$ .