If $a,b,c$ are the sides of a triangle $ABC$ such that $x^3-24x^2+180x-420=0. $Then $ \tan A+\tan B+\tan C$ is
Try: $a+b+c=24$
$ab+bc+ca=180$ and $abc=420$
And use $a=k\sin A,B=k\sin B, c=k\sin c$ where $k=2R$ circumradius of circle.
After substituting $a,b,c$ in equation. Not getting the term $\tan A+\tan B+\tan C$
Could some help me thanks
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If the sides $a,b,c$ of a triangle are the roots of the cubic, the coefficients can be expressed in terms of three linear measures of the triangle, inradius $r$, circumradius $R$ and semiperimeter $\rho=\tfrac12(a+b+c)$ as follows:
\begin{align} x^3-2\rho\,x^2+(\rho^2+r^2+4\,r\,R)\,x-4\,r\rho\,R . \end{align}
Hence, for the given equation \begin{align} x^3-24\,x^2+180\,x-420 &=0 \end{align}
we have
\begin{align} \rho&=12 ,\\ 4\,r\,R&=35 ,\\ r&=\sqrt{180-12^2-35}=1 ,\\ R&=\tfrac{35}4 . \end{align}
The sum of the tangents can be also expressed in similar way as
\begin{align} \tan\alpha+\tan\beta+\tan\gamma &=\frac{2\,r\,\rho}{\rho^2-(r+2\,R)^2} \end{align}
With known values of $r,R$ and $\rho$ it is trivial to find that \begin{align} \tan\alpha+\tan\beta+\tan\gamma&=-\frac{96}{793} . \end{align}
hint: $S = \dfrac{abc}{4R}= \dfrac{420}{4R} = \dfrac{105}{R}\implies \dfrac{ab\sin C}{2} = \dfrac{105}{R}\implies \sin C= \dfrac{210}{abR}= \dfrac{210c}{abcR}= \dfrac{c}{2R}$. Use Heron formula to solve for $S = \sqrt{p(p-a)(p-b)(p-c)},p = 12$. Simplify the expression inside square root, and you can solve for $S$ then $R$. To solve for the cosines, solve the system: $a = b\cos C + c\cos B, b = a\cos C + c\cos A, c = a\cos B + b\cos A$ for $\cos A, \cos B, \cos C$ in terms of $a,b,c$. After that, substitute, say $\tan A = \dfrac{\sin A}{\cos A}= \dfrac{ma}{f(a,b,c)}$ with $m$ is a constant and $f(a,b,c)$ is a symmetric polynomial of degree max $3$ of $a,b,c$ and can be found by the above info or at this point, the above info can be useful to get the answer.