Posted by **ken** on Wednesday, June 9, 2010 at 8:55am.

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

- geometry -
**Reiny**, Wednesday, June 9, 2010 at 9:11am
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

- geometry -
**drwls**, Wednesday, June 9, 2010 at 9:16am
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

- geometry -
**drwls**, Wednesday, June 9, 2010 at 9:21am
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

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