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Q.
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
TS EAMCET 2020
Solution:
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 $A B C$
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R$
$\{$where $2 R \rightarrow$ circumradius $\}$
$\therefore a=2 R \sin A, b=2 R \sin B, c=2 R \sin C$
Substituting the values of $a, b, c$ in Eq . (i), we get
$\{(2 R \sin A)^{3}+(2 R \sin B)^{3}+(2 R \sin C)^{3} \{\sin^{3} A+ \sin ^{3}B+ \sin^{3} C\}=8$
or $\frac{(2 R)^{3}\left(\sin ^{3} A+\sin ^{3} B+\sin ^{3} C\right)}{\sin ^{3} A+\sin ^{3} B+\sin ^{3} C}=8$
or $(2 R)^{3}=8$ or $2 R=2$