geometry
posted by ken .
In triangle ABC, BC=40m , AB=50m and Angle A=53 degree. Determine the area of a triangle.

find angle C first of all
sinC/50 = sin53/40
sinC=.99829
angle C = 86.65° or 93.35° (we have the ambiguous case, you can actually draw two triangles satisfying your information.
then angle B = 40.35° or 33.65°
Case 1
AB=50
BC=40
angle B = 40.35
area = (1/2)(50)(40)sin40.35
= 647.41
Do the second case in the same way, you should get 554.16 
According to the law of sines,
40/sin 53 = 50/sin C
Therefore
sin C = (50/40)*sin53 = 0.998
C = 86.7 degrees
B = 180  53  86.7 = 40.3 degrees
With all sides and all angles, you can solve for the area. If you pick AC as the base,
Area = (AC)*(1/2)*(BC)*sin 86.7
= 50*(0.5)*40*.998 = 998 m^2 
Since my numbers don't agree with Reiny's, check mine for mistakes. Reiny is right about there being two cases. I assumed all acute angles