Question

The radii of described circle of $△ABC$ are $r_1,r_2$ and $r_3$ respectively (opposite to vertices $A,B$ and $C$). If $r_2+r_3=2R$ and $r_1+r_2=3R$ then

• A. $\angle A={90}^o$
• B. $\angle B={45}^o$
• C. $\angle C={60}^o$
• D. $\angle B={90}^o$
  

Answer 1

Answer: (A), (C)

The correct options are
A $\angle A={90}^o$
C $\angle C={60}^o$

In triangle ABC

$r_1=\frac{△}{s-a}$,$r_2=\frac{△}{s-b}$,$r_3=\frac{△}{s-a}$,$R=\frac{abc}{4△}$

so $r_2+r_3=2R$, put values

$\frac{△}{s-b}+\frac{△}{s-c}=\frac{2abc}{4△}$

$\frac{{△}^2(2s-b-c)}{(s-b)(s-c)}=\frac{abc}{2}$

put $s=\frac{a+b+c}{2},△=\sqrt{s(s-a)(s-b)(s-c)}$

than we get $\left(\frac{a+b+c}{2}\right)\left(\frac{-a+b+c}{2}\right)=\frac{bc}{2}$

$b^2+c^2-a^2=0$

so cosA=0

Angle A=$\frac{\pi }{2}$

so similarily in $r_1+r^2=3R$

we will get $\frac{a^2+b^2-c^2}{2bc}=\frac{1}{2}$

so cosC=$\frac{1}{2}$

Angle C= $\frac{\pi }{3}$