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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 . 

Grade:12

1 Answers

Nirabhra Agrawal
33 Points
11 years ago



 

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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