A) 1
B) 2
C) 3
D) 4
Correct Answer: D
Solution :
[d] We have \[\tan A=\frac{a}{2{{d}_{1}}};\] |
\[{{d}_{1}}=R\cos A\] etc. |
Similarly \[\tan B=\frac{b}{2{{d}_{2}}}\] |
and \[\tan C=\frac{C}{2{{d}_{3}}}\] |
In \[\Delta ABC,\tan A+\tan B+\tan C\] |
\[=\,\,\,\,\tan A\cdot \tan B\cdot \tan C\] |
\[\Rightarrow \frac{a}{2{{d}_{1}}}+\frac{b}{2{{d}_{2}}}+\frac{c}{2{{d}_{3}}}=\frac{abc}{8{{d}_{1}}{{d}_{2}}{{d}_{3}}}\] |
\[\therefore \,\,\,\,\,\,\,4\left( \frac{a}{{{d}_{1}}}+\frac{b}{{{d}_{2}}}+\frac{c}{{{d}_{3}}} \right)=\frac{abc}{{{d}_{1}}{{d}_{2}}{{d}_{3}}}\] |
\[\Rightarrow \lambda =4\] |
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