In a triangle ABC the median AD and the altitude AM divides the angle A in three equal parts.Show that (cos A/3.sin 2 A/3) = 3a 3 /128R .

Question

Nirabhra Agrawal · Answer

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.

(3)

(cos A/3.sin²A/3) = 3a³ b^2/128R c^2.

Comments from shri Sagar Singh is awaited.

Thanks

G P Agrawal.

f/o Nirabhra AGRAWAL

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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.sin 2 A/3) = 3a 3 /128R .

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 .

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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.sin 2 A/3) = 3a 3 /128R .

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 .

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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 .