Q66 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Mathematics

In a triangle $A B C$, 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
  • Correct answer
    C
    2
  • D
    6

Solution

Given that, $\frac{a^3+b^3+c^3}{\sin ^3 A+\sin ^3 B+\sin ^3 C}=8 \quad \ldots$ (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 $\frac{(2 R \sin A)^3+(2 R \sin B)^3+(2 R \sin C)^3}{\sin ^3 A+\sin ^3 B+\sin ^3 C}=8$ or $\quad \frac{(2 R)^3\left(\sin ^3 A+\sin ^3 B+\sin ^3 C\right)}{\sin ^3 A+\sin ^3 B+\sin ^3 C}=8$ or $\quad(2 R)^3=8$ or $2 R=2$

Hence, the correct answer is option: 'C'.

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