Answer
The angles of the triangle are as follows:
$A = 63^{\circ}50', B = 43^{\circ}30',$ and $C = 72^{\circ}40'$
The lengths of the sides are as follows:
$a = 327~ft, b = 251~ft,$ and $c = 348~ft$
Work Step by Step
We can use the law of cosines to find $c$:
$c^2 = a^2+b^2-2ab~cos~C$
$c = \sqrt{a^2+b^2-2ab~cos~C}$
$c = \sqrt{(327~ft)^2+(251~ft)^2-(2)(327~ft)(251~ft)~cos~72^{\circ}40'}$
$c = \sqrt{121024.46~ft^2}$
$c = 348~ft$
We can use the law of cosines to find $B$:
$b^2 = a^2+c^2-2ac~cos~B$
$2ac~cos~B = a^2+c^2-b^2$
$cos~B = \frac{a^2+c^2-b^2}{2ac}$
$B = arccos(\frac{a^2+c^2-b^2}{2ac})$
$B = arccos(\frac{(327)^2+(348)^2-(251)^2}{(2)(327)(348)})$
$B = arccos(0.725)$
$B = 43^{\circ}30'$
We can find angle $A$:
$A+B+C = 180^{\circ}$
$A = 180^{\circ}-B-C$
$A = 180^{\circ}-43^{\circ}30'-72^{\circ}40'$
$A = 63^{\circ}50'$
