In triangle ABC,if BC is the hypotenuse,then | vedclass

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(B) Since $BC$ is the hypotenuse,the triangle is right-angled at $A$,therefore $\angle A = \frac{\pi}{2}$.We know that $r_2 = \frac{\Delta}{s-b}$ and

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$a$

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(B) Since $BC$ is the hypotenuse,the triangle is right-angled at $A$,therefore $\angle A = \frac{\pi}{2}$.We know that $r_2 = \frac{\Delta}{s-b}$ and $r_3 = \frac{\Delta}{s-c}$.Thus,$r_2 + r_3 = \Delta \left( \frac{1}{s-b} + \frac{1}{s-c} \right) = \Delta \left( \frac{s-c+s-b}{(s-b)(s-c)} \right) = \Delta \left( \frac{2s-b-c}{(s-b)(s-c)} \right)$.Since $2s = a+b+c$,we have $2s-b-c = a$.Also,$\Delta^2 = s(s-a)(s-b)(s-c)$,so $(s-b)(s-c) = \frac{\Delta^2}{s(s-a)}$.Substituting these,$r_2 + r_3 = \Delta \left( \frac{a}{\Delta^2 / s(s-a)} \right) = \frac{as(s-a)}{\Delta}$.Since $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$,we have $\frac{s(s-a)}{\Delta} = \cot \frac{A}{2}$.Therefore,$r_2 + r_3 = a \cot \frac{A}{2} = a \cot \frac{\pi}{4} = a(1) = a$.

In triangle $ABC$,if $BC$ is the hypotenuse,then $r_2 + r_3 =$

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