SOLUTION: The three angles of a triangle are 60 degrees, 76.67 degrees, 43.33 degrees respectively. If the area of the triangle is 88.33 square units find the length of the longest side of t

Question 1076896: The three angles of a triangle are 60 degrees, 76.67 degrees, 43.33 degrees respectively. If the area of the triangle is 88.33 square units find the length of the longest side of the triangle. What is the perimeter of the triangle?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The three angles of a triangle are 60 degrees, 76.67 degrees, 43.33 degrees respectively. If the area of the triangle is 88.33 square units find the length of the longest side of the triangle. What is the perimeter of the triangle?
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Note:: The side opposite the 76.67 degree angle is the longest.
Let that side be "c" ; let "a" be opposite 60 deg ; let "b" be opp 43.33
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Formula: Area = a*b*sin(C)
Solve for "c"::
88.33 = a*c*sin(B) = 0.6862 ac
88.33 = b*c*sin(A) = 0.8660 bc
88.33 = a*b*sin(C) = 0.9731 ab
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88.33/0.6862a = 88.33/0.8660b
128.72/a = 102/b
b = 102a/128.72
b = 0.79a
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So, 88.33 = 0.9731a(0.79a)
88.33 = 0.7711a^2
a = sqrt(114.55) = 10.7
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Then c = 88.33/(0.6862*10.7) = 12.03
And b = 88.33/(0.9731*10.7) = 8.48
etc.
Cheers,
Stan H.
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