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

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) (Show Source): You can put this solution on YOUR website!
By hypothesis, .
Since the triangle also obeys the Sine Law, we also have .

Now and ===> and
By adding corresponding sides of the last two equations, we get

, which implies that .

Similarly, and ===> .

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

Hence its area is square units.
RELATED QUESTIONS
In ΔABC, Given (a/cosA)=(b/cosB)=(c/cosC) . Prove that ΔABC is an equilateral... (answered by Edwin McCravy)

If A, B, C are angles of a triangle, prove that cosA + cosB + cosC =... (answered by MathLover1)

if a/cosb=b/cosa then prove that triangle abc is right angled... (answered by solver91311)

in triangle ABC, a=2, b=3, and c=4. What is the value of... (answered by stanbon)

In triangle ABC, a=8, b=9, and cosC= 2/3. Find... (answered by ikleyn)

in triangle ABC, a = 7, b = 5, and c = 8. What is the value of... (answered by Alan3354)

For any triangle ABC, we have cosA + cosB + cosC =1+4SinA/2SinB/2SinC/2 and SinA/2 +... (answered by robertb)

If cos(A-B)+cos(B-C)+cos(C-A)= -3/2, prove that:... (answered by robertb)

cosA+cosB+cosC=3/2 ,then show that the triangle is equilateral... (answered by Edwin McCravy)