If the sides `a, b` & `c` `/_\ABC` is such that `a/(1+alpha^(2)beta^(2))=b/(alpha^(2)+beta^(2))=c/((1-alpha^(2))(1+beta^(2)))` then

Options

A. `A=2tan^(-1)(alpha//beta)`
B. `B=2tan^(1-)(alpha beta)`
C. area of `/_\ABC=(alpha beta ab)/(alpha^(2)+beta^(2))`
D. area of `/_\ABC=(alpha beta bc)/(alpha^(2)+beta^(2))`

Answer

A, B, D

Explanation

`a/(1+alpha^(2)beta^(2))=b/(alpha^(2)+beta^(2))=c/((1-alpha^(2))(1+beta^(2)))=(2s)/(2+2beta^(2))`
`implies s/(1+beta^(2))=a/(1+alpha^(2)beta^(2))implies(s-a)/(beta^(2)(1-alpha^(2)))=a/(1+alpha^(2)beta^(2))`
`implies s=a=(a(1-alpha^(2))beta^(2))/(1+alpha^(2)beta^(2))`
Similarly `s-b=(b(1-alpha^(2)))/(alpha^(2)+beta^(2))`
`s-c=(calpha^(2))/(1-alpha^(2))` & `s=(b(1+beta^(2)))/(alpha^(2)+beta^(2))`
`:. tan^(2)(A//2)=((s-b)(s-c))/(s(s-a))=((b(1-alpha^(2)))/(alpha^(2)+beta^(2)).(calpha^(2))/(1-alpha^(2)))/(b(1+beta^(2))/(alpha^(2)+beta^(2)) (a(1-alpha^(2))beta^(2))/(1+alpha^(2)beta^(2)))`
`[Qa=(c(1+alpha^(2)beta^(2)))/((1-alpha^(2))(1+beta^(2)))]`
`implies (alpha^(2))/(beta^(2))`
`A=2tan^(-1)(alpha//beta)`
`B=2tan^(-1)(alphabeta)`
`:.` Reqd area `=1/2 bc sin A = 1/2 bc. (2tan (A//2))/(1+tan^(2)(A//2))=(bcalphabeta)/(alpha^(2)+beta^(2))`