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Acute triangle, Trigonometric function! (Posted on 2007-10-13) Difficulty: 3 of 5
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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Question re: Solution | Comment 6 of 9 |
(In reply to Solution by K Sengupta)

In going from

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


Or,  2R < R*(sin A + sin B + sin C) (in terms of the law of sines)

it would seem that a substitution has been made of R*sinA for (1/2)*AB, etc.  It is not the case that they are equal, but apparently the inference is valid as R*sinA is bigger than (1/2)*AB, etc.  I don't see, though what application of the law of sines you are using.

Could you make the application of the law of sines explicit, in terms of side1/sin(angle1) = side2/sin(angle2), and, in light of the inequality, rather than equality in the substitution, demonstrate that a larger term has been substituted for a smaller rather than the other way around?

Geometer's Sketchpad confirms that the substituted terms are indeed bigger, but I'd like to see how this is proved.

  Posted by Charlie on 2007-10-14 10:37:43
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