Question

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

• A. $\frac{1}{2}$
• B. $2$
• C. $8$
• D. $64$

Answer 1

The correct option is B $2$

By Sine Rule,
$a=2R\sin A,b=2R\sin B,c=2R\sin C$
where $k=2R$
Substituting $a=k\sin A,b=k\sin B,c=k\sin C$ in
$\frac{a^3+b^3+c^3}{{\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C}=8$
$\Rightarrow \frac{{\left(k\sin A\right)}^3+{\left(k\sin A\right)}^3+{\left(k\sin A\right)}^3}{{\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C}=8$
$\Rightarrow \frac{k^3({\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C)}{{\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C}=8$
$\Rightarrow k^3=8$
$\therefore k=2$
Hence, $a=k\sin A=2\sin A\le 2$