Question

If $\frac{r}{r_2}=\frac{r_1}{r_3}$ then where r is radius of in circle

• A. $A={90}^0$
• B. $B={90}^0$
• C. $C={90}^0$
• D. $C={60}^o$

Answer 1

The correct option is C $C={90}^0$

$\text{FORMULA USED IN SOLVING PROBLEM}$

1. $s=\frac{a+b+c}{2}$
   
2. $r=\frac{\Delta }{s}$ (∆ = area of triangle)
   
3. $r_1=\frac{\Delta }{s-a},r_2=\frac{\Delta }{s-b},r_3=\frac{\Delta }{s-c}$

$\frac{\frac{\Delta }{s}}{\frac{\Delta }{s-b}}=\frac{\frac{\Delta }{s-a}}{\frac{\Delta }{s-c}}$....(using formulas of $r,r_1,r_2,r_3$)

$\frac{s-b}{s}=\frac{s-c}{s-a}$

$s^2-sa-sb+ab=s^2-sc$

$ab=s(a+b-c)$

$2ab=(a+b+c)(a+b-c)$

$2ab=(a+b{)}^2-c^2$

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

$\therefore \angle c=\pi /2$.