Let f, g, h be the lengths of the perpendicul | vedclass

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(A) Let $R$ be the circumradius of $\Delta ABC$. The perpendicular distance from the circumcentre to the side $a$ is $f = R \cos A$.Similarly,$g = R \

Vedclass · Accepted Answer

$1/4$

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(A) Let $R$ be the circumradius of $\Delta ABC$. The perpendicular distance from the circumcentre to the side $a$ is $f = R \cos A$.Similarly,$g = R \cos B$ and $h = R \cos C$.We know that $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.Thus,$\frac{a}{f} = \frac{2R \sin A}{R \cos A} = 2 	an A$.Similarly,$\frac{b}{g} = 2 	an B$ and $\frac{c}{h} = 2 	an C$.Substituting these into the given equation: $2 	an A + 2 	an B + 2 	an C = \lambda \frac{(2R \sin A)(2R \sin B)(2R \sin C)}{(R \cos A)(R \cos B)(R \cos C)}$.$2(	an A + 	an B + 	an C) = \lambda \cdot 8 	an A 	an B 	an C$.In any triangle,$	an A + 	an B + 	an C = 	an A 	an B 	an C$.Therefore,$2(	an A 	an B 	an C) = 8 \lambda (	an A 	an B 	an C)$.$2 = 8 \lambda \Rightarrow \lambda = 1/4$.

Let $f, g, h$ be the lengths of the perpendiculars from the circumcentre of the $\Delta ABC$ on the sides $a, b$ and $c$ respectively. If $\frac{a}{f} + \frac{b}{g} + \frac{c}{h} = \lambda \frac{abc}{fgh}$,then the value of $\lambda$ is:

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