Chapter 6, Problem 2CT

To determine

Whether the Law of Sines or the Law of Cosines 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 2CT

Thesolution is $A={42}^0,B={110}^0,C={28}^0,a=15.6,b=21.90,c=10.94$

Explanation of Solution

Given info:

  $B={110}^0,C={28}^0,a=15.6$

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 $B={110}^0,C={28}^0,a=15.6$ , We have

  $\begin{array}{l}B={110}^0,C={28}^0,a=15.6\\ A={180}^0-{110}^0-{28}^0={42}^0\\ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\\ \frac{15.6}{\sin {42}^0}=\frac{b}{\sin {110}^0}=\frac{c}{\sin {28}^0}\\ b=23.31\times \sin {110}^0=21.90\\ c=23.31\times \sin {28}^0=10.94\\ A={42}^0,B={110}^0,C={28}^0,a=15.6,b=21.90,c=10.94\end{array}$

Conclusion:

Thus,the solution is $A={42}^0,B={110}^0,C={28}^0,a=15.6,b=21.90,c=10.94$