Question

In $\Delta ABC$, $a\le b\le c$, if $\frac{a^3+b^3+c^3}{si{n}^{3}A+si{n}^{3}B+si{n}^{3}C}=8$, 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,

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=K$

$\Rightarrow a=K\sin A,b=K\sin B$ and $c=K\sin C$

Putting the values of $a,b$ and $c$ in the given equation, we get,

$\frac{K^3{\sin }^{3}A+K^3{\sin }^{3}B+K^3{\sin }^{3}C}{{\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C}=8$

$\Rightarrow K^3=8$

$\Rightarrow K=2$

$\Rightarrow a=K\sin A=2\sin A\le 2$