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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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 Solution | Comment 5 of 9 |

Each of the angles ABC (or, B) , BCA(or, C) and CAB(or, A) of the triangle ABC is acute (given).  Accordingly, the circumcenter (E, say)  must lie inside the triangle.

Let X, Y and Z respectively denote the mid points of   BC, CA and AB.
Accordingly, it follows that E must be located in the interior of one of  the quadrilaterals BCYZ, CAZX, or  ABXY.

Without loss of generality, we can assume that E is located in the interior of BCYZ.

Let R be the length of the circumradius of the triangle ABC.

Then,
2R =  BE + EC

Since triangles AZY and ABC are similar, with BZ = AB/2, and
AY = AC/2, we must have ZY = BC/2.

Now, it can easily be shown that:

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)

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

Edited on October 14, 2007, 6:25 am
 Posted by K Sengupta on 2007-10-14 06:08:14

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