Determine whether angle abc has no solution or one solution. then solve the triangle. a = 112, a = 7, b = 4

 
There is only one triangle possible

Angles: 
A = 112, B = 31.9932, C = 36.0068

Sides:
a = 7, b = 4, c = 4.4384

See the attached image for a visual diagram.
==============================================

Explanation:

Use the law of sines to solve for angle B
sin(A)/a = sin(B)/b
sin(112)/7 = sin(B)/4
0.132454836366684 = sin(B)/4
4*0.132454836366684 = 4*sin(B)/4
0.529819345466736 = sin(B)
sin(B) = 0.529819345466736
arcsin(sin(B)) = arcsin(0.529819345466736)
B = arcsin(0.529819345466736) ... or ... B = 180 - arcsin(0.529819345466736)
B = 31.9932495421859 ... or ... B = 148.006750457814
B = 31.9932 ... or ... B = 148.0068

If B = 31.9932, then C = 180-A-B = 180-112-31.9932 = 36.0068

If B = 148.0068, then C = 180-A-B = 180-112-148.0068 = -80.0068
A negative angle is not possible, so we toss out this value

So only angle B = 31.9932 is possible

--------------------------

Once we know the measure of angle C, we can use that to find side c by using the law of cosines
c^2 = a^2 + b^2 - 2*a*b*cos(C)
c^2 = (7)^2 + (4)^2 - 2*(7)*(4)*cos(36.0068)
c^2 = 49 + 16 - 56*0.808947228919362
c^2 = 49 + 16 - 45.3010448194843
c^2 = 19.6989551805157
c = sqrt(19.6989551805157)
c = 4.4383505022154
c = 4.4384

By this point, we're done since we found the remaining missing angles and missing side.

Ugucycucucuvuvjvxyxtxu