With usual notation in a Delta ABC,if R = k f | vedclass

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(C) We know that $r_1 + r_2 = \frac{\Delta}{s-a} + \frac{\Delta}{s-b} = \frac{\Delta(s-b+s-a)}{(s-a)(s-b)} = \frac{\Delta c}{(s-a)(s-b)} = \frac{c}{r}

Vedclass · Accepted Answer

$1/4$

Vedclass · Answer

(C) We know that $r_1 + r_2 = \frac{\Delta}{s-a} + \frac{\Delta}{s-b} = \frac{\Delta(s-b+s-a)}{(s-a)(s-b)} = \frac{\Delta c}{(s-a)(s-b)} = \frac{c}{r} \cdot \frac{\Delta^2}{s(s-a)(s-b)(s-c)} \cdot \frac{s(s-c)}{\Delta} = \frac{c}{r} \cdot \frac{s(s-c)}{\Delta} = \frac{c}{r} \cdot \frac{s(s-c)R}{abc/4R} = \dots$ Alternatively,using $r_1 = s 	an(A/2)$,$r_2 = s 	an(B/2)$,$r_3 = s 	an(C/2)$,we have $r_1 + r_2 = \frac{c}{\cos(C/2)} \cos(A/2) \cos(B/2) / \dots = \frac{4R \cos(C/2) \cos(A/2) \cos(B/2)}{\sin(C/2)} = \frac{4R \cos(C/2) \sin(C/2)}{\sin(C/2)} = 4R \cos^2(C/2)$ is not correct. Correct identity: $r_1 + r_2 = \frac{c}{\sin C} \cos(A/2) \cos(B/2) \dots = 4R \cos(C/2) \cos(A/2) \cos(B/2) / \sin(C/2) = 4R \cos(C/2) \sin(C/2) / \sin(C/2) = 4R \cos^2(C/2)$ is wrong. Actually,$r_1 + r_2 = \frac{c \cos(C/2)}{\sin(A/2) \sin(B/2)}$. Using $r_1 = \frac{\Delta}{s-a}$,the expression simplifies to $R = k(4R)$. Thus,$k = 1/4$.

With usual notation in a $\Delta ABC$,if $R = k \frac{(r_1 + r_2)(r_2 + r_3)(r_3 + r_1)}{r_1 r_2 + r_2 r_3 + r_3 r_1}$,where $k$ has the value equal to:

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