Chapter 6, Problem 27CLT

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 27CLT

The solution is $A={30}^0,B={60}^0,C={90}^0,a=5.77,b=10,c=11.547$

Explanation of Solution

Given info:

  $A={30}^0,C={90}^0,b=10$

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 $A={30}^0,C={90}^0,b=10$ , We have

  $\begin{array}{l}A={30}^0,C={90}^0,b=10\\ B={180}^0-{30}^0-{90}^0={60}^0\\ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\\ \frac{a}{\sin {30}^0}=\frac{10}{\sin {60}^0}=\frac{c}{\sin {90}^0}\\ \frac{a}{\sin {30}^0}=\frac{c}{\sin {90}^0}=11.547\\ a=5.77\\ c=11.547\\ A={30}^0,B={60}^0,C={90}^0,a=5.77,b=10,c=11.547\end{array}$

Conclusion:

Thus,the solution is $A={30}^0,B={60}^0,C={90}^0,a=5.77,b=10,c=11.547$