In triangle ABC,if r_1+r_2=3 R and r_2+r_3=2 | vedclass

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(C) Given $r_1+r_2=3 R$ and $r_2+r_3=2 R$. Using the formula $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c}$ and $R =

Vedclass · Accepted Answer

$A=90^{\circ}, a 
eq b 
eq c$

Vedclass · Answer

(C) Given $r_1+r_2=3 R$ and $r_2+r_3=2 R$. Using the formula $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,$r_3 = \frac{\Delta}{s-c}$ and $R = \frac{abc}{4\Delta}$. For $r_1+r_2=3 R$: $\frac{\Delta}{s-a} + \frac{\Delta}{s-b} = 3R$ $\Rightarrow \frac{\Delta(2s-a-b)}{(s-a)(s-b)} = 3R$ $\Rightarrow \frac{\Delta c}{(s-a)(s-b)} = 3R$. Since $\Delta^2 = s(s-a)(s-b)(s-c)$,we get $\frac{s(s-c)}{c} = 3R$ $\Rightarrow \cos^2 \frac{C}{2} = \frac{3}{4}$ $\Rightarrow \angle C = 60^{\circ}$. For $r_2+r_3=2 R$: $\frac{\Delta}{s-b} + \frac{\Delta}{s-c} = 2R \Rightarrow \frac{\Delta a}{(s-b)(s-c)} = 2R$. This leads to $\cos^2 \frac{A}{2} = \frac{1}{2} \Rightarrow \angle A = 90^{\circ}$. Since $\angle A + \angle B + \angle C = 180^{\circ}$,we have $90^{\circ} + \angle B + 60^{\circ} = 180^{\circ} \Rightarrow \angle B = 30^{\circ}$. Thus,$\angle A = 90^{\circ}, \angle B = 30^{\circ}, \angle C = 60^{\circ}$. The sides ratio is $a:b:c = \sin 90^{\circ} : \sin 30^{\circ} : \sin 60^{\circ} = 1 : \frac{1}{2} : \frac{\sqrt{3}}{2} = 2 : 1 : \sqrt{3}$.

In $\triangle ABC$,if $r_1+r_2=3 R$ and $r_2+r_3=2 R$,then

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