Question

If the sine of the angles of triangle $\Delta ABC$ satisfy the equations ${{c}^{3}}{{x}^{3}}-{{c}^{2}}\left( a+b+c \right){{x}^{2}}+lx+m=0$(where $a,b,c$ are the length of the sides of the triangle $\Delta ABC$), then $\Delta ABC$ is \[\]
1. always right angled for any $l,m$\[\]
2. right angled only when $l=c\left( ab+bc+ca \right)=c\sum{ab,m=-abc}$\[\]
3. right angled when \[l=\dfrac{c\sum{ab}}{4},m=\dfrac{-abc}{8}\]
4. never right angled \[\]

Answer 1

Hint: Use sum of the roots formula of a polynomial equation to prove that the triangle is right-angled. Then use the product of the roots formula to find out the value of $m$.\[\]

Complete step-by-step solution:
We know that in any cubic polynomial equation ${{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+{{a}_{3}}{{x}^{3}}=0$, the sum of the roots is given by the negative ratio of the coefficient of second highest power to coefficient highest power. Let the three roots be ${{\alpha }_{1}},{{\alpha }_{2}},{{\alpha }_{3}}$ then in symbols sum of roots is ${{\alpha }_{1}}+{{\alpha }_{2}}+{{\alpha }_{3}}=-\dfrac{{{a}_{2}}}{{{a}_{3}}}$. Similarly the products of the roots are given the negative ratio of the constant term to the coefficient highest power. In symbols, product of the roots is $ {{\alpha }_{1}}{{\alpha }_{2}}{{\alpha }_{3}}=\dfrac{-{{a}_{0}}}{{{a}_{3}}} $. Also we use the formula of $\sum{{{\alpha }_{1}}{{\alpha }_{2}}=}{{\alpha }_{1}}{{\alpha }_{2}}+{{\alpha }_{2}}{{\alpha }_{3}}+{{\alpha }_{2}}{{\alpha }_{3}}=\dfrac{-{{a}_{1}}}{{{a}_{3}}}$\[\]
The given polynomial equation is \[\]
${{c}^{3}}{{x}^{3}}-{{c}^{2}}\left( a+b+c \right){{x}^{2}}+lx+m=0.....(1)$\[\]
As given in the question the sine of the angles of triangle $\Delta ABC$ which are
$\sin A,\sin B,\sin C$ satisfy equation (1).So sum of the roots,\[\]
$\sin A+\sin B+\sin C=-\left( \dfrac{-{{c}^{2}}(a+b+c)}{{{c}^{3}}} \right)=\dfrac{a+b+c}{c}.....\left( 2 \right)$ \[\]
From the law of the sine in any triangle,\[\]
\[\begin{align}
  & \dfrac{a}{sinA}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R\left( \text{say} \right) \\
 & \Rightarrow \sin A=\dfrac{a}{2R},\sin B=\dfrac{a}{2R},\sin C=\dfrac{c}{2R} \\
\end{align}\]
Putting above values in equation (2)
\[\begin{align}
  & \dfrac{a}{2R}+\dfrac{b}{2R}+\dfrac{c}{2R}=\dfrac{a+b+c}{c} \\
 & \Rightarrow \dfrac{a+b+c}{c}=\dfrac{a+b+c}{2R} \\
 & \Rightarrow c=2R \\
 & \Rightarrow 2R\sin C=2R(\text{from sine law}) \\
 & \Rightarrow \sin C=1\Rightarrow C=\dfrac{\pi }{2} \\
\end{align}\]
So the triangle $\Delta ABC$ is right angled.
Now the product of the roots is from equation(2)
\[\begin{align}
  & \sin A\sin B\sin C=\dfrac{-m}{{{c}^{3}}} \\
 & \Rightarrow \dfrac{abc}{8{{R}^{3}}}=\dfrac{-m}{{{c}^{3}}} \\
 & \Rightarrow m=\dfrac{-\left( abc \right){{c}^{3}}}{8{{R}^{3}}}.=-abc{{\left( \dfrac{c}{2R} \right)}^{3}}=-abc{{\left( \sin C \right)}^{3}}=-abc...(3) \\
\end{align}\]
Using the formula of sum product of two roots for polynomial equation(2),
\[\begin{align}
  & \sum{\sin A\sin B}=\dfrac{l}{{{c}^{3}}} \\
 & \Rightarrow \sum{\dfrac{ab}{{{R}^{2}}}}=\dfrac{l}{{{c}^{3}}} \\
 & \Rightarrow l=\dfrac{{{c}^{3}}}{4{{R}^{2}}}\sum{ab} \\
 & \Rightarrow l={{\left( \dfrac{c}{2R} \right)}^{2}}\left( c\sum{ab} \right)=\sin C\left( c\sum{ab} \right) \\
\end{align}\]
Putting $ \sin C=1 $ in above calculation
\[l=c\sum{ab}=c(ab+bc+ca)....(4)\]
From equation (3) and (4) the right angle has the property $m=-abc$ and $l=c(ab+bc+ca)$. Hence the correct option is B.

Note: We can similarly find the sum and product of roots of polynomial equation of ${{n}^{\text{th}}}$ degree say \[{{a}_{n}}{{x}^{n}}+{{a}_{n-1}}{{x}^{n-1}}+...+{{a}_{1}}x+{{a}_{o}}=0\] where ${{a}_{n}},{{a}_{n-1}},...,{{a}_{0}}$ real coefficients. Here the sum of the roots is $-\dfrac{{{a}_{n-1}}}{{{a}_{n}}}$ and the product of the roots is ${{\left( -1 \right)}^{n}}\dfrac{{{a}_{0}}}{{{a}_{n}}}$. It is called Vieta’s formula.