If the area`(!)` and an angle`(theta)` of a triangle are given , when the side opposite to the given angle is minimum , then the length of the remaining two sides are

Options

A. `sqrt((2!)/(sintheta)),sqrt((3!)/(sintheta))`
B. `sqrt((2!)/(sintheta)),sqrt((2!)/(sintheta))`
C. `sqrt((4!)/(sintheta)),sqrt((4!)/(sintheta))`
D. `sqrt((6!)/(sintheta)),sqrt((6!)/(sintheta))`

Explanation

Let `theta` be the angle opposite to side c(=AB) of `DeltaABC`. Then,
`c^(2)=a^(2)+b^(2)-2abcosthetarArrc^(2)=(a-b)^(2)+2b(1-costheta)` ltbr. Also,
`Delta=1/2absinthetarArr2ab=(4Delta)/(sintheta)` ..(i)
`thereforec^(2)=(a-b)^(2)+4Delta((1-costheta)/(sintheta))`
`rArrc^(2)=(a-b)^(2)+4Deltatan(theta)/2`
`rArr`c is minimum when a -b = 0 i.e. a =b.
Putting a =b in (i), we get `a=sqrt((2Delta)/(sintheta))=b`