A) \[R\]
B) \[2R\]
C) \[3R\]
D) \[4R\]
Correct Answer: D
Solution :
Given, \[{{r}_{1}}+{{r}_{3}}=k\,{{\cos }^{2}}\,B/2\] ie, \[s\,\,\tan \,A/2+s\,\tan C/2=k\,{{\cos }^{2}}B/2\] \[\Rightarrow \] \[s\left[ \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}+\sqrt{\frac{(s-a)(s-b)}{s(s-c)}} \right]\] \[=k\frac{s(s-b)}{ac}\] \[\Rightarrow \] \[k=\frac{ac}{s-b}\left[ \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}+\sqrt{\frac{(s-a)\,(s-b)}{s(s-c)}} \right]\] \[=\frac{ac}{s-b}\times \sqrt{\frac{s-b}{\sqrt{s}}}\left[ \frac{\sqrt{(s-c)}}{\sqrt{(s-a)}}+\frac{\sqrt{(s-a)}}{\sqrt{(s-c)}} \right]\] \[=\frac{ac}{\sqrt{s(s-b)}}\left[ \frac{s-c+s-a}{\sqrt{s-a}\sqrt{s-c}} \right]\] \[=\frac{ac(2s-a-c)}{\sqrt{s(s-a)(s-b)(s-c)}}\] \[=\frac{ac}{\Delta }(a+b+c-a-c)=\frac{abc}{\Delta }\] \[\Rightarrow \] \[k=4R\]
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