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If in a $\triangle A B C, r_1 < r_2 < r_3$, then:
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Verified Answer
The correct answer is:
$a < b < c$
We have
$r_1=\frac{\Delta}{s-a}, r_2=\frac{\Delta}{s-b}, r_3=\frac{\Delta}{s-c}$
Since, $r_1 < r_2 < r_3$
$\Rightarrow \quad \frac{\Delta}{s-a} < \frac{\Delta}{s-b} < \frac{\Delta}{s-c}$
$\Rightarrow \quad(s-a)>(s-b)>(s-c)$
$\Rightarrow \quad-a>-b>-c$
$\therefore \quad a < b < c$
$r_1=\frac{\Delta}{s-a}, r_2=\frac{\Delta}{s-b}, r_3=\frac{\Delta}{s-c}$
Since, $r_1 < r_2 < r_3$
$\Rightarrow \quad \frac{\Delta}{s-a} < \frac{\Delta}{s-b} < \frac{\Delta}{s-c}$
$\Rightarrow \quad(s-a)>(s-b)>(s-c)$
$\Rightarrow \quad-a>-b>-c$
$\therefore \quad a < b < c$
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