Chapter 6, Problem 33RE

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 33RE

The solution is $A={57.79}^0,B={89.63}^0,C={32.58}^0,a=11,b=13,c=7$

Explanation of Solution

Given info:

  $a=13,b=15,c=24$

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 $a=13,b=15,c=24$ ,

We have

  $\begin{array}{l}a=13,b=15,c=24\\ \cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{{15}^2+{24}^2-{13}^2}{2\times 15\times 24}\\ A={28.62}^0\\ \cos B=\frac{a^2+c^2-b^2}{2ac}=\frac{{13}^2+{24}^2-{15}^2}{2\times 13\times 24}\\ B={33.56}^0\\ \cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{{13}^2+{15}^2-{24}^2}{2\times 13\times 15}\\ C={117.82}^0\\ A={28.62}^0,B={33.56}^0,C={117.82}^0,a=13,b=15,c=24\end{array}$

Conclusion:

Thus,the solution is $A={28.62}^0,B={33.56}^0,C={117.82}^0,a=13,b=15,c=24$