Answer
The angles of the triangle are as follows:
$A = 53.1^{\circ}, B = 81.3^{\circ},$ and $C = 45.6^{\circ}$
The lengths of the sides are as follows:
$a = 7.23~m, b = 8.94~m,$ and $c = 6.46~m$
Work Step by Step
$a = 7.23~m$
$b = 8.94~m$
$C = 45.6^{\circ}$
We can use the law of cosines to find $c$:
$c^2 = a^2+b^2-2ab~cos~C$
$c = \sqrt{a^2+b^2-2ab~cos~C}$
$c = \sqrt{(7.23~m)^2+(8.94~m)^2-(2)(7.23~m)(8.94~m)~cos~45.6^{\circ}}$
$c = \sqrt{41.749~m^2}$
$c = 6.46~m$
We can use the law of sines to find $A$:
$\frac{c}{sin~C} = \frac{a}{sin~A}$
$sin~A = \frac{a~sin~C}{c}$
$A = arcsin(\frac{a~sin~C}{c})$
$A = arcsin(\frac{7.23~sin(45.6^{\circ})}{6.46})$
$A = arcsin(0.799634)$
$A = 53.1^{\circ}$
We can find angle $B$:
$A+B+C = 180^{\circ}$
$B = 180^{\circ}-A-C$
$B = 180^{\circ}-53.1^{\circ}-45.6^{\circ}$
$B = 81.3^{\circ}$
