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Let P Q R be a triangle of area with a=...

Let `P Q R` be a triangle of area `` with `a=2,b=7/2,a n dc=5/2, w h e r ea , b ,a n dc` are the lengths of the sides of the triangle opposite to the angles at `P ,Q ,a n dR` respectively. Then `(2sinP-sin2P)/(2sinP+sin2P)e q u a l s` `3/(4)` (b) `(45)/(4)` (c) `(3/(4))^2` (d) `((45)/(4))^2`

A

`(3)/(4Delta)`

B

`(45)/(4Delta)`

C

`((3)/(4Delta))^(2)`

D

`((45)/(4Delta))^(2)`

Text Solution

Verified by Experts

The correct Answer is:
C

`(2 sin P - 2 sin P cos P)/(2 sin P + 2 sin P cos P)`
`= (1- cos P)/(1 +cos P) = (2"sin"^(2) (P)/(2))/(2 "cos"^(2) (P)/(2)) = "tan"^(2) (P)/(2)`
`= ((s-b) (s-c))/(s(s-a))`
`= (((s-b) (s-c))^(2))/(Delta^(2)) = ((((1)/(2)) ((3)/(2)))^(2))/(Delta^(2)) = ((3)/(4Delta))^(2)`
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