Question

In a triangle $ABC$, if $a<b<c$ and $\frac{a^3+b^3+c^3}{{\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C}=8$, then the maximum value of $c$ is

• A. 3
• B. 4
• C. 2
• D. 6

Answer 1

Answer: (C)

Given that, $\frac{a^3+b^3+c^3}{{\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C}=8$ ...(i)
Using sine rule for a triangle $ABC$
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$
$\{$where $2R\to$ circumradius $\}$
$\therefore a=2R\sin A,b=2R\sin B,c=2R\sin C$
Substituting the values of $a,b,c$ in Eq . (i), we get
$\{(2R\sin A{)}^3+(2R\sin B{)}^3+(2R\sin C{)}^3\{{\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C\}=8$
or $\frac{(2R{)}^3\left({\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C\right)}{{\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C}=8$
or $(2R{)}^3=8$ or $2R=2$