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In a triangle ABC the median AD and the altitude AM divides the angle A in three equal parts.Show that (cos A/3.sin2A/3) = 3a3 /128R .
Today at 5.00 AM I solved this problem with two more results as follows:
(cos A/3.sin2A/3) = 3a3 b^2/128R c^2.
in which b^2/c^2 = 3.
Hence this may be a good question for questions having more than one correct answers. correct options will be as follows
(1) (cos A/3.sin2A/3) = 3a3 /128R .
(2)
b^2/c^2 = 3.
Comments from shri Sagar Singh is awaited.
Thanks
G P Agrawal.
f/o Nirabhra AGRAWAL
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