In triangle ABC,if 4r_1 = 5r_2 = 6r_3,then si | vedclass

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(B) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.Given $4r_1 = 5r_2 = 6r_3 = \lambda$.Then $s-a =

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$\frac{25}{33}$

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(B) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.Given $4r_1 = 5r_2 = 6r_3 = \lambda$.Then $s-a = \frac{\lambda}{4}$,$s-b = \frac{\lambda}{5}$,and $s-c = \frac{\lambda}{6}$.Summing these: $(s-a) + (s-b) + (s-c) = 3s - (a+b+c) = 3s - 2s = s$.So,$s = \lambda(\frac{1}{4} + \frac{1}{5} + \frac{1}{6}) = \lambda(\frac{15+12+10}{60}) = \frac{37\lambda}{60}$.Then $a = s - (s-a) = \lambda(\frac{37}{60} - \frac{15}{60}) = \frac{22\lambda}{60}$,$b = \lambda(\frac{37}{60} - \frac{12}{60}) = \frac{25\lambda}{60}$,$c = \lambda(\frac{37}{60} - \frac{10}{60}) = \frac{27\lambda}{60}$.Using $\sin^2 \frac{A}{2} = \frac{(s-b)(s-c)}{bc}$,$\sin^2 \frac{B}{2} = \frac{(s-a)(s-c)}{ac}$,$\sin^2 \frac{C}{2} = \frac{(s-a)(s-b)}{ab}$.Sum $= \frac{(\lambda/5)(\lambda/6)}{(25\lambda/60)(27\lambda/60)} + \frac{(\lambda/4)(\lambda/6)}{(22\lambda/60)(27\lambda/60)} + \frac{(\lambda/4)(\lambda/5)}{(22\lambda/60)(25\lambda/60)}$.Sum $= \frac{3600}{30 	imes 675} + \frac{3600}{24 	imes 594} + \frac{3600}{20 	imes 550} = \frac{120}{675} + \frac{150}{594} + \frac{180}{550} = \frac{8}{45} + \frac{25}{99} + \frac{18}{55} = \frac{88 + 125 + 162}{495} = \frac{375}{495} = \frac{25}{33}$.

In $\triangle ABC$,if $4r_1 = 5r_2 = 6r_3$,then $\sin^2 \frac{A}{2} + \sin^2 \frac{B}{2} + \sin^2 \frac{C}{2} =$

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