Question 172762
1. Find the area of triangle ABC if A=56 degrees, b=20 ft, and c=12 ft. Round to the nearest tenth.
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If you sketch it, then draw the altitude from the end of b, that altitude = 10*sin(56) and it's the h of A = bh/2.
Area = 12*20*sin(56)/2
Area = 99.5 sq units
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2. In triangle ABC, A=35 degrees, a=43 and c=20. Determine whether triangle ABC has no solution, one solution, or two solutions. Then solve the triangle. Round to the nearest tenth.
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Use the Law of Sines
a/sin(A) = c/sin(C)
43/sin(35) = 20/sin(C)
sin(C) = 20*sin(35)/43 = 0.26677974
C = 15.4727 degs or 164.5273 degs (from inverse sine)
C cannot be 164... degs, tho, because adding A would exceed 180 degs.
So C = 15.4727 degs, and there is only one solution.
Angle B = 180 - (A+C) = 129.5273 degs
side b/sin(B) = a/sin(A)
b = 43*sin(B)/sin(A)
b = 57.825