In a `Delta ABC,` show that ` 2R^(2) sin A sin B sin C=Delta.`

Explanation

Let `N_(1)` be the amount of parent atom of A at any instant `t-Delta t`.Therefore, ` N_(1)=N_(1)^(0) e^(-lambda_(A)(t-Deltat)`…..(1),Where `N_(1)^(0) ` is the number of parent atom of A at t=0.The number of daughter atoms at anytime t is given by `N_(2)=(lambda _(A)N_(1)^(0))/(lambda_(B)-lambda_(A))[e^(-lambda_(A)t)-e^(-lambda_(B)t)]`.......(2),For transient atom `lambda _(A) < lambda_(B)`, hence `e^(-lambda_(B))` , t term can be neglected in (2) `therefore N_(2)-(lambda _(A)N_(1)^(0))/(lambda_(B)-lambda_(A))e^(-lambda_(A)t)`........(3)Given that activity of A at `(t-Delta t) =activity of B at t.`implies` `lambda _(A) N_(1)^(0) e^(-lambda _(A)(t-Delta t))=lambda_(B) (lambda_(A) N_(1)^(0) )/(lambda_(B)-lambda _(A))e^(-lambda_(A)t)`,`implies` `e^(-lambda _(A) Delta t)=(lambda_(B))/(lambda _(B)-lambda_(A))` `implies` `Delta t=(1)/(lambda_(A))In [(lambda_(B))/(lambda_(B)-lambda_(A))]`. The avergae life of A `iota =(1)/(lambda_(A))` and that of B is `iota =(1)/(lambda_(B)) `,`therefore` `Delta t =iota _(A)` In `[(1)/((1-lambda_(B))/(lambda_(A)))]`` implies` ` Delta t cong iota _(A).(iota_(B))/(iota_(A))` as `iota _(B)/iota _(A) rArr 0` (given) `implies` `Delta t= iota _(B)`