Question

In a triangle $ABC$, if $\frac{a^3+b^3+c^3}{{\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C}=343$, the diameter of the circle circumscribing the triangle is

• A. $14$ units
• B. $7$ units
• C. $21$ units
• D. $noneofthese$

Answer 1

Answer: (B)

$7$ units

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

Using sine rule

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\to$ circumradius

$a=2R\sin A$

$b=2R\sin B$

$c=2R\sin C$

substitute $a,b,c$ in given equation

$8A^3\left(\frac{{\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C}{{\sin }^{3}A+{\sin }^{3}B+{\sin }^{3}C}\right)=243$

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

$2R=7$

Diameter of circumcircle $=7$