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Let PQR be a triangle of area Delta with...

Let PQR be a triangle of area `Delta` with `a=2,b=7/2 and c=5/2`, where a,b and c are the lengths of the sides of the triangle opposite to the angles at P,Q and R respectively. Then `((2sinP-sin2P)/ (2sinP+sin2P))` equals

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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