Chapter 6.2, Problem 31E

To determine

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

Answer to Problem 31E

Thesolution is $A={16.05}^0,B={40}^0,C={123.95}^0,a=5,b=11.62,c=15$

Explanation of Solution

Given info:

  $a=5,c=15,B={40}^0$

Formula used:

  

Law of sines

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

Law of cosines

  $\begin{array}{l}\cos A=\frac{b^2+c^2-a^2}{2bc}\\ \cos B=\frac{a^2+c^2-b^2}{2ac}\\ \cos C=\frac{a^2+b^2-c^2}{2ab}\end{array}$

Calculation:

Given that $a=5,c=15,B={40}^0$ , We have

  $\begin{array}{l}a=5,c=15,B={40}^0\\ \cos B=\frac{a^2+c^2-b^2}{2ac}\\ \cos {40}^0=\frac{5^2+{15}^2-b^2}{2\times 5\times 15}\\ b=11.62\\ \cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{{11.62}^2+{15}^2-5^2}{2\times 11.62\times 15}\\ A={16.05}^0\\ \cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{{11.62}^2+5^2-{15}^2}{2\times 11.62\times 5}\\ C={123.95}^0\\ A={16.05}^0,B={40}^0,C={123.95}^0,a=5,b=11.62,c=15\end{array}$

Conclusion:

Thus,the solution is $A={16.05}^0,B={40}^0,C={123.95}^0,a=5,b=11.62,c=15$