Chapter 6.1, Problem 20E

To determine

The solution of triangle using law of sines given that $B=2^{0}45\text{'},b=6.2,c=5.8$ .

Answer to Problem 20E

Thesolution is $A=\text{174}{\text{.68}}^0,B=2^{0}45\text{'},C=\text{2}{\text{.57}}^0,a=\text{12}\text{.02},b=\text{6}\text{.2},c=\text{5}\text{.8}$

Explanation of Solution

Formula used:

Given info:

  $B=2^{0}45\text{'},b=6.2,c=5.8$

Formula used:

  

Law of sines

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

Calculation:

Given that $B=2^{0}45\text{'},b=6.2,c=5.8$ ,

We have

  $\begin{array}{l}\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\\ \frac{a}{\sin A}=\frac{6.2}{\sin 2^{0}45\text{'}}=\frac{5.8}{\sin C}\\ \frac{a}{\sin A}=\frac{5.8}{\sin C}=129.23\\ C=\text{2}{\text{.57}}^0\\ A={180}^0-(\text{2}{\text{.75}}^0+\text{2}{\text{.57}}^0)=\text{174}{\text{.68}}^0\\ \frac{a}{\sin \text{174}{\text{.68}}^0}=129.23\\ a=0.093\times 129.23=\text{12}\text{.02}\\ A=\text{174}{\text{.68}}^0,B=2^{0}45\text{'},C=\text{2}{\text{.57}}^0,a=\text{12}\text{.02},b=\text{6}\text{.2},c=\text{5}\text{.8}\end{array}$

Conclusion:

Thus,the solution is $A=\text{174}{\text{.68}}^0,B=2^{0}45\text{'},C=\text{2}{\text{.57}}^0,a=\text{12}\text{.02},b=\text{6}\text{.2},c=\text{5}\text{.8}$ .