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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))^(5)`

Text Solution

Verified by Experts

The correct Answer is:
C
If `DeltaABC` has sides a,b,c,

Then `tan(A//2)=sqrt(((s-b)(s-a))/(s(s-a)))`
Where `s=(a+b+c)/(2)`
`rArr s=(2+(7)/(2)+(5)/(2))/(2)=4`
`:. (2 sinP-sin2P)/(2sin P+sin2P)=(2sinP(1-cosP))/(2sinP(1+cosP))`
`=(2sin^(2)(P//2))/(2cos^(2)(P//2))=tan^(2)(P//2)`

`((s-b)(s-c))/(s(s-a))xx((s-b)(s-c))/((s-b)(s-c))`
`=([(s-b)^(2)(s-c)^(2)])/(Delta^(2))((4-(7)/(2))^(2)(4-(5)/(2))^(2))/(Delta^(2))=((3)/(4Delta))^(2)`
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