Answer
There are two possible solutions for this triangle.
The angles of the triangle are as follows:
$A = 83^{\circ}, B = 58^{\circ}30',$ and $C = 38^{\circ}30'$
The lengths of the sides are as follows:
$a = 1252~in, b = 1075~in,$ and $c = 785~in$
There is another possible solution for this triangle:
The angles of the triangle are as follows:
$A = 20^{\circ}, B = 121^{\circ}30',$ and $C = 38^{\circ}30'$
The lengths of the sides are as follows:
$a = 431~in, b = 1075~in,$ and $c = 785~in$
Work Step by Step
$C = 38^{\circ}30' = 38.5^{\circ}$
$b = 1075~in$
$c = 785~in$
We can use the law of sines to find angle $B$:
$\frac{c}{sin~C} = \frac{b}{sin~B}$
$sin~B = \frac{b~sin~C}{c}$
$B = arcsin(\frac{b~sin~C}{c})$
$B = arcsin(\frac{1075~sin~38.5^{\circ}}{785})$
$B = arcsin(0.8525)$
$B = 58.5^{\circ} = 58^{\circ}30'$
We can find angle $A$:
$A+B+C = 180^{\circ}$
$A = 180^{\circ}-B-C$
$A = 180^{\circ}-58^{\circ}30'-38^{\circ}30'$
$A = 83^{\circ}$
We can use the law of sines to find the length of side $a$:
$\frac{a}{sin~A} = \frac{c}{sin~C}$
$a = \frac{c~sin~A}{sin~C}$
$a = \frac{(785~in)~sin~(83^{\circ})}{sin~38.5^{\circ}}$
$a = 1252~in$
Note that there is another possibility for angle $B$:
$B = 180^{\circ}- 58.5^{\circ} = 121.5^{\circ} = 121^{\circ}30'$
We can find angle $A$:
$A+B+C = 180^{\circ}$
$A = 180^{\circ}-B-C$
$A = 180^{\circ}-121^{\circ}30'-38^{\circ}30'$
$A = 20^{\circ}$
We can use the law of sines to find the length of side $a$:
$\frac{a}{sin~A} = \frac{c}{sin~C}$
$a = \frac{c~sin~A}{sin~C}$
$a = \frac{(785~in)~sin~(20^{\circ})}{sin~38.5^{\circ}}$
$a = 431~in$
