prove that (cosA*cosB*cosC)<=1/8
Asked by | 14th Dec, 2009, 12:11: PM
Expert Answer:
A, B,C are the angles of the triangle then A+B+C= 180o
If we consider the triangle to be acute angled and each angle to be 60o then cos A + cos B+CosC=3/2
In general cos A + cos B+CosC
3/2
use A.M G.M
(cos A +CosB+cosC)/3 (cosA.cosB.cosC)^1/3
so, cosA.cosB.cosC
1/8
Answered by | 16th Dec, 2009, 09:14: AM
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