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+CosC3/2
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
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Sign Up
Verify mobile number
Enter the OTP sent to your number
Change