SOLUTION: In ΔABC, Given (a/cosA)=(b/cosB)=(c/cosC) . Prove that ΔABC is an equilateral triangle.

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Question 252352: In ΔABC, Given (a/cosA)=(b/cosB)=(c/cosC) . Prove that ΔABC is an equilateral triangle.
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
In ΔABC, Given (a/cosA)=(b/cosB)=(c/cosC) . Prove that ΔABC is an equilateral triangle.
a%2F%28cosA%29=b%2F%28cosB%29=c%2F%28cosC%29

By the law of sines:

a%2F%28sinA%29=b%2F%28sinB%29=c%2F%28sinC%29

and therefore

%28sinA%29%2Fa=%28sinB%29%2Fb=%28sinC%29%2Fc

Multiplying equals by equals:





%28%28sinA%29%2F%28cosA%29%29=%28%28sinB%29%2F%28cosB%29%29=%28%28sinC%29%2F%28cosC%29%29

tanA=tanB=tanC

A,B, and C are angles of a triangle and thus are less than 180°.

Any two angles less than 180° which have the same tangent must
be equal in measure.  Thus

A = B = C

Thus triangle ABC is equiangular and every equiangular
triangle is also equilateral.

Edwin