Chapter 6, Problem 22CR

To determine

To calculate:

Solve the triangle using law of cosine.

Answer to Problem 22CR

Triangle using law of cosine is $B=85.84°,A=75.06°,19.10°$ .

Explanation of Solution

Given information:

  $a=6.2,b=6.4,c=2.1$

Calculation:

Let’s,

  $a=6.2,b=6.4,c=2.1$

For solving the triangle by using law of cosine, first need to angle opposite to the determine longest side hat means side $b$ ,

  $\cos B=\frac{c^2+a^2-b^2}{bc}$

Put the values in above equation,

  $\begin{array}{l}\cos B=\frac{{\left(6.2\right)}^2+{\left(2.1\right)}^2-{\left(6.4\right)}^2}{2\left(6.2\right)\left(2.1\right)}\\ \cos B=\frac{37.2+4.41-40.96}{26.04}\frac{37.2+4.41-40.96}{26.04}\\ \cos B=\frac{0.65}{26.04}\frac{0.65}{6.04}\\ \cos B=0.02496\\ B=85.84°\end{array}$

To determine $A$ ,use sine law,

  $\begin{array}{l}\sin A=a\left(\frac{\sin B}{b}\right)\\ \sin A=6.2\left(\frac{\sin \left(85.84°\right)}{6.4}\right)6.2\left(\frac{\sin \left(85.84°\right)}{6.4}\right)\\ \sin A=6.2\frac{0.0999731}{6.4}\frac{0.99731}{6.4}\\ \sin A=6.2\left(0.15582\right)\\ A=75.06°\end{array}$

To find the third angle of the triangle is,

  $C=180°-\left(A+B\right)$

Substitute for $A$ and $C$ ,

  $\begin{array}{l}C=180°-75.06°-85.84°\\ C=19.10°\end{array}$

Graph of the triangle is as below:

  

Therefore, triangle using law of cosine is $B=85.84°,A=75.06°,19.10°$ .