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Nirabhra Agrawal Grade: 12
        

 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 . 

6 years ago

Answers : (1)

Nirabhra Agrawal
33 Points
										



































 


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






6 years ago
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