Question

In a triangle $ABC$, if $\frac{a^3+b^3+c^3}{{\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C}=343$

then the diameter of the circle circumscribing the triangle is?

• A. $7$ units
• B. $14$ units
• C. $21$ units
• D. None of the above

Answer 1

Answer: (A)

$7$ units

Given, $\frac{a^3+b^3+c^3}{{\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C}=343$

$\Rightarrow \frac{8R^3({\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C)}{{\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C}=343$ [by sine rule]

$\Rightarrow (2R{)}^3=7^3$

$\therefore 2R=7$