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In a triangle ABC if the sides a,b,c are in AP and 2/1!.9!+2/3!.7!+1/5!.5!=8 a /(2b)!,then the maximum value of tanA.tanB is equal to A)1/2 B)1/3 C) 1/4 D) 1/5

In a triangle ABC if the sides a,b,c are in AP and 2/1!.9!+2/3!.7!+1/5!.5!=8a/(2b)!,then the maximum value of tanA.tanB is equal to


 


A)1/2     B)1/3   C) 1/4  D) 1/5


 

Grade:11

1 Answers

Ramesh V
70 Points
11 years ago

2/1!.9! + 2/3!.7! + 1/5!.5! = 23a/(2b)!

simply solving above gives

1/(4*6!) *(1/5 +13/63)

512/10! = 23a/(2b)!

so a=3 , b =5 , c=7

area of triangle =15*31/2/4

sin A = 2*area /bc = 3*31/2/14

sin B = 2*area /ac = 5*31/2/14

Cos A = 13/14  , cos B = 11/14

tan A.TanB = 45/143

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regards

Ramesh

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