Question

The radii of the escribed circles of $△ABC$ are $r_a,r_b$ and $r_c$ respectively. If $r_a+r_b=3R$ and $r_b+r_c=2R$, then the smallest angle of the triangle is

• A. ${\tan }^{-1}(\sqrt{2}-1)$
• B. $\frac{1}{2}{\tan }^{-1}(\sqrt{3})$
• C. $\frac{1}{2}{\tan }^{-1}(\sqrt{2}+1)$
• D. ${\tan }^{-1}(2-\sqrt{3})$

Answer 1

Answer: (B)

We have $r_a+r_b=3R\Rightarrow \frac{\Delta }{s-a}+\frac{\Delta }{s-b}=3R=\frac{3abc}{4\Delta }\left(R=\frac{abc}{4\Delta }\right)$
$\Rightarrow \frac{\Delta (s-b+s-a)}{(s-a)(s-b)}=\frac{3abc}{4\Delta }$
$\Rightarrow \frac{c\Delta }{(s-a)(s-b)}=\frac{3abc}{4\Delta }$
$\Rightarrow \frac{{\Delta }^2}{(s-a)(s-b)}=\frac{3ab}{4}$
$\Rightarrow 4s(s-c)=3ab\Rightarrow (a+b+c)(a+b-c)=3ab$
$\Rightarrow (a+b{)}^2-c^2=3ab\Rightarrow a^2+b^2-c^2=ab$
$\Rightarrow c^2=a^2+b^2-ab$
$\Rightarrow a^2+b^2-2ab\cos C=a^2+b^2-ab\left(As{c}^2=a^2+b^2-2ab\cos C\right)$
$\Rightarrow \cos C=\frac{1}{2}\Rightarrow \angle C=60^{\circ }\dots .(1)$
$|||ly$ from $r_b+r_c=2R$
$\Rightarrow \frac{\Delta }{s-b}+\frac{\Delta }{s-c}=2R\Rightarrow \frac{\Delta (2s-b-c)}{(s-b)(s-c)}=\frac{2abc}{4\Delta }\Rightarrow \frac{2{\Delta }^2}{(s-b)(s-c)}=bc$
$\Rightarrow 2s(s-a)=bc\Rightarrow (b+c+a)(b+c-a)=2bc\Rightarrow (b+c{)}^2-a^2=2bc$
$\Rightarrow b^2+c^2=a^2\Rightarrow \angle A=90^{\circ }\Rightarrow \angle B=30^{\circ }$