Chapter 6, Problem 26CLT

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 place.

Answer to Problem 26CLT

The solution is $A={30}^0,B={52.45}^0,C={97.55}^0,a=5.04,b=8,c=10$

Explanation of Solution

Given info:

  $A={30}^0,b=8,c=10$

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 $A={30}^0,b=8,c=10$ , We have

  $\begin{array}{l}A={30}^0,b=8,c=10\\ \cos A=\frac{b^2+c^2-a^2}{2bc}\\ \cos {30}^0=\frac{8^2+{10}^2-a^2}{2\times 8\times 10}\\ a=5.04\\ \cos B=\frac{a^2+c^2-b^2}{2ac}=\frac{{5.04}^2+{10}^2-8^2}{2\times 5.04\times 10}\\ B={52.45}^0\\ \cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{{5.04}^2+8^2-{10}^2}{2\times 5.04\times 8}\\ C={97.55}^0\\ A={30}^0,B={52.45}^0,C={97.55}^0,a=5.04,b=8,c=10\end{array}$

Conclusion:

Thus,the solution is $A={30}^0,B={52.45}^0,C={97.55}^0,a=5.04,b=8,c=10$