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a^2+b^2-2ab*cos(α)= b^2+c^2-2bc*cos(α) or c^2-a^2 = 2b(c-a)cos(α) or cos(α) = (a+c)/(2b).Since α>0. cos(α) < 1 and thus (a+c)/(2b) < 1 which gives us a + c < 2bVihaan
http://geometri-problemleri.blogspot.com/2010/07/problem-81-ve-cozumu.html
a^2+b^2-2ab*cos(α)= b^2+c^2-2bc*cos(α) or c^2-a^2 = 2b(c-a)cos(α) or cos(α) = (a+c)/(2b).
ReplyDeleteSince α>0. cos(α) < 1 and thus (a+c)/(2b) < 1 which gives us a + c < 2b
Vihaan
http://geometri-problemleri.blogspot.com/2010/07/problem-81-ve-cozumu.html
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