Question

In a triangle ABC, a $\ge$ b $\ge$ c. If $\frac{a3+b^3+c^3}{si{n}^{3}A+si{n}^{3}B+si{n}^{3}C}=8$,

then the maximum value of $a$ is?

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

Answer 1

Answer: (B)

$2$

according to sine rule,

$a^3+b^3+c^3=8R^3(si{n}^{3}A+si{n}^{3}B+si{n}^{3}C)...................\left(1\right)$

Given:

$a^3+b^3+c^3=8(si{n}^{3}A+si{n}^{3}B+si{n}^{3}C).....................\left(2\right)$

Comparing $\left(1\right)$ and $\left(2\right)$

we get $R=1$

Now, again from sine Rule

$a=2RsinA\Rightarrow a=2sinA$

we know that maximum value of $sinA$ is $1$

so maximum value of $2sinA=2$

Therefore, Answer is $B$