Question

If the sine of the angles of $△ABC$ satisfy the equation $c^3{x}^3-c^2(a+b+c)x^2+lx+m=0$ (where $a,b,c$ are the sides of $△ABC$), then $△ABC$ is

• A. always right angled for any $l,m$
• B. right angled only when $l=c(ab+bc+ca)=c\sum ab,m=-abc$
• C. right angled only when $l=\frac{c\sum ab}{4},m=\frac{-abc}{8}$
• D. never right angled

Answer 1

The correct option is B right angled only when $l=c(ab+bc+ca)=c\sum ab,m=-abc$

$\sin A,\sin B,\sin C$ are the roots of the equation
$c^3{x}^3-c^2(a+b+c)x^2+lx+m=0$ (where $a,b,c$ are the sides of $△ABC$)
$\therefore \sin A+\sin B+\sin C=\frac{c^2(a+b+c)}{c^3}=\frac{a+b+c}{c}$
or $\frac{a}{2R}+\frac{b}{2R}+\frac{c}{2R}=\frac{a+b+c}{c}$
$\Rightarrow 2R=c$ or $c=2R$ or $a+b+c\ne 0$
$\Rightarrow 2R\sin C=2R$ (using sine rule)
$\therefore \sin C=1\Rightarrow \angle C=\frac{\pi }{2}$
$\Rightarrow$ Triangle is a right angled triangle.
and $\sum \sin A\sin B=\frac{l}{c^3}$
and product of the roots$=\sin A\sin B\sin C=\frac{-m}{c^3}$
Using sine rule,
$\sum \frac{a}{2R}.\frac{b}{2R}=\frac{l}{c^3}$ and $\frac{a}{2R}\frac{b}{2R}\frac{c}{2R}=\frac{-m}{c^3}$
$\Rightarrow \sum \frac{ab}{R^2}=\frac{l}{c^3}$ and $\frac{abc}{8R^3}=\frac{-m}{c^3}$
$\Rightarrow l=\frac{c^3}{4R^2}\sum ab$ and $m=\frac{-\left(abc\right)c^3}{8R^3}$
$\Rightarrow l={\left(\frac{c}{2R}\right)}^{2}c\sum ab=\sin C.c\sum ab$
We know that $\angle C=\frac{\pi }{2}\therefore \sin C=1$
Hence, $l=c\sum ab=c(ab+bc+ca)$
and $m=-\left(abc\right){\left(\frac{c}{2R}\right)}^3$
$=-\left(abc\right){\left(\sin C\right)}^3$
$=-abc$ ,since $\sin C=1$