Answer
The angles of the triangle are as follows:
$A = 112.8^{\circ}, B = 21.7^{\circ},$ and $C = 45.5^{\circ}$
The lengths of the sides are as follows:
$a = 15.7~m, b = 6.28~m,$ and $c = 12.2~m$
Work Step by Step
We can use the law of cosines to find $a$:
$a^2 = b^2+c^2-2bc~cos~A$
$a = \sqrt{b^2+c^2-2bc~cos~A}$
$a = \sqrt{(6.28~m)^2+(12.2~m)^2-(2)(6.28~m)(12.2~m)~cos~112.8^{\circ}}$
$a = \sqrt{247.658~m^2}$
$a = 15.7~m$
We can use the law of cosines to find $B$:
$b^2 = a^2+c^2-2ac~cos~B$
$2ac~cos~B = a^2+c^2-b^2$
$cos~B = \frac{a^2+c^2-b^2}{2ac}$
$B = arccos(\frac{a^2+c^2-b^2}{2ac})$
$B = arccos(\frac{(15.7)^2+(12.2)^2-(6.28)^2}{(2)(15.7)(12.2)})$
$B = arccos(0.929)$
$B = 21.7^{\circ}$
We can find angle $C$:
$A+B+C = 180^{\circ}$
$C = 180^{\circ}-A-B$
$C = 180^{\circ}-112.8^{\circ}-21.7^{\circ}$
$C = 45.5^{\circ}$
