Question

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))^(5)`

Answer 1

Correct Answer - 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)`