Question

In ΔABC prove that `sin  "A"/(2). sin  "B"/(2). sin  "C"/(2) = ["A(ΔABC)"]^2/"abcs"`

Answer 1

L.H.S.

= `sin  "A"/(2). sin  "B"/(2). sin  "C"/(2)` 

`= sqrt(((s - b)(s - c))/"bc") .sqrt(((s - a)(s - c))/"ac") . sqrt(((s -a )(s - b))/"ab"`

`= sqrt(((s - a)^2(s -  b)^2(s - c)^2)/(a^2b^2c^2)`

`= ((s - a)(s - b)(s - c))/"abc"`

`= (s(s - a)(s - b)(s - c))/"abcs"`

`= (["A(ΔABC")]^2/"abcs"   ...[∵ "A(ΔABC") = sqrt(s(s - a)(s - b)(s - c))]`
= R.H.S.