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 Acute triangle, Trigonometric function! (Posted on 2007-10-13)
Prove that in an acute triangle sin(A) + sin(B) + sin(C) > 2

 See The Solution Submitted by Chesca Ciprian Rating: 4.0000 (1 votes)

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 re(2): Solution | Comment 8 of 9 |
(In reply to re: Solution by Charlie)

After the step:

BE + EC < BZ + ZY + YC = (1/2)*(AB + BC + AC).......(*)

But, BE = EC = R, where R is the length of circumradius of the triangle ABC, while E is the circumcenter of the triangle.

Hence, BE+ EC = 2R........(i)

In terms of the law of sines,
(Reference: http://en.wikipedia.org/wiki/Law_of_sines), we have:

BC/sin A = CA/sin B = AB/sin C = 2R, giving:

(1/2)*(AB + BC + AC) = R(sin C + sin A + sin B).......(ii)

Substituting (i) and (ii) in (*), we have:

2R< R(sin A + sin B + sin C)

Or, sin A + sin B + sin C > 2

Edited on October 14, 2007, 3:13 pm
 Posted by K Sengupta on 2007-10-14 10:59:29

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