Question

The radii of the escribed circles of $\triangle ABC$ are $r _{ a }, r _{ b }$ and $r _{ c }$ respectively. If $r_a+r_b=3 R$ and $r_b+r_c=2 R$, 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 $r_a+r_b=3 R \Rightarrow \frac{\Delta}{s-a}+\frac{\Delta}{s-b}=3 R=\frac{3 a b c}{4 \Delta} \left(R=\frac{a b c}{4 \Delta}\right)$
$\Rightarrow \frac{\Delta(s-b+s-a)}{(s-a)(s-b)}=\frac{3 a b c}{4 \Delta} $
$ \Rightarrow \frac{c \Delta}{(s-a)(s-b)}=\frac{3 a b c}{4 \Delta} $
$\Rightarrow \frac{\Delta^2}{(s-a)(s-b)}=\frac{3 a b}{4}$
$\Rightarrow 4 s(s-c)=3 a b \Rightarrow (a+b+c)(a+b-c)=3 a b $
$\Rightarrow (a+b)^2-c^2=3 a b \Rightarrow a^2+b^2-c^2=a b $
$\Rightarrow c^2=a^2+b^2-a b$
$\Rightarrow a^2+b^2-2 a b \cos C=a^2+b^2-a b \left(A s c^2=a^2+b^2-2 a b \cos C\right) $
$\Rightarrow \cos C=\frac{1}{2} \Rightarrow \angle C=60^{\circ} \ldots .(1) $
$|||\,ly$ from $r_b+r_c=2 R$
$\Rightarrow \frac{\Delta}{ s - b }+\frac{\Delta}{ s - c }=2 R \Rightarrow \frac{\Delta(2 s - b - c )}{( s - b )( s - c )}=\frac{2 abc }{4 \Delta} \Rightarrow \frac{2 \Delta^2}{( s - b )( s - c )}= bc$
$\Rightarrow 2 s(s-a)=b c \Rightarrow (b+c+a)(b+c-a)=2 b c \Rightarrow (b+c)^2-a^2=2 b c $
$\Rightarrow b^2+c^2=a^2 \Rightarrow \angle A=90^{\circ} \Rightarrow \angle B=30^{\circ}$