Question 500711: What is the perimeter of a triangle with side a: 2y, angle A: 2z-y, angle B: x+y-10 and angle C: 3x-z+20?
Answer by cleomenius(959)   (Show Source):
You can put this solution on YOUR website! You could use the law of sin to find each side.
It would be set up as follows:
2y/(2z -y) = Side B /sin(x + y + 10) = Side C/ sin( 3x - z + 20)
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To find side B:
2y/(2z -y) = Side B /sin(x + y + 10) =
2y * (sin(x + y - 10))/ sin(2z - y).
This would be side b, the best we are able to determine it from the information given.
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To find side C:
2y/(2z -y) = Side C/sin( 3x - z + 20) =
2y * (sin(3x -z+20))/ sin(2z - y)
This would be side c, the best we are able to determine it from the information given.
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To find our perimeter, we would add the three.
2y + 2y * (sin(x + y - 10))/ sin(2z - y) + 2y * (sin(3x -z+20))/ sin(2z - y)
The only thing left we can do is factor out a 2y.
(2y) (1 + (sin(x + y - 10))/ sin(2z - y) + (sin(3x -z+20))/ sin(2z - y))
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Cleomenius.
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