Chapter 6, Problem 25RE

To determine

The solution of triangle using law of cosines given that $C={43}^0,a=22.5,b=31.4$ .

Answer to Problem 25RE

The solution is $A={45.76}^0,B={91.24}^0,C={93}^0,a=22.5,b=31.4,c=21.42$

Explanation of Solution

Given info:

  $C={43}^0,a=22.5,b=31.4$

Formula used:

  

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 $C={43}^0,a=22.5,b=31.4$ , We have

  $\begin{array}{l}C={43}^0,a=22.5,b=31.4\\ \cos C=\frac{a^2+b^2-c^2}{2ab}\\ \cos {43}^0=\frac{{22.5}^2+{31.4}^2-c^2}{2\times 22.5\times 31.4}\\ c=21.42\\ \cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{{31.4}^2+{21.42}^2-{22.5}^2}{2\times 31.4\times 21.42}\\ A={45.76}^0\\ \cos B=\frac{a^2+c^2-b^2}{2ac}=\frac{{22.5}^2+{21.42}^2-{31.4}^2}{2\times 22.5\times 21.42}\\ B={91.24}^0\\ A={45.76}^0,B={91.24}^0,C={93}^0,a=22.5,b=31.4,c=21.42\end{array}$

Conclusion:

Thus,the solution is $A={45.76}^0,B={91.24}^0,C={93}^0,a=22.5,b=31.4,c=21.42$