SOLUTION: In a triangle if CosA/a = CosB/b = CosC/c and if a = 2 then find the area of triangle ABC?

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Question 1184155: In a triangle if CosA/a = CosB/b = CosC/c and if a = 2 then find the area of triangle ABC?
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
By hypothesis, CosA%2Fa+=+CosB%2Fb+=+CosC%2Fc.
Since the triangle also obeys the Sine Law, we also have sinA%2Fa+=+SinB%2Fb+=+SinC%2Fc.

Now CosA%2Fa+=+CosB%2Fb and sinA%2Fa+=+SinB%2Fb ===> Cos%5E2A%2Fa%5E2+=+Cos%5E2B%2Fb%5E2 and Sin%5E2A%2Fa%5E2+=+Sin%5E2B%2Fb%5E2
By adding corresponding sides of the last two equations, we get

1%2Fa%5E2+=+1%2Fb%5E2, which implies that a+=+b.

Similarly, CosA%2Fa+=+CosC%2Fc and sinA%2Fa+=+SinC%2Fc ===> a+=+c.

Therefore, the triangle is actually an isosceles triangle, each side having a measure of a = 2 units.

Hence its area is square units.