Question

In a $△ABC$, if $a$ is the arithmetic mean and b & c $(b\ne c)$ are two geometric means between any two positive real numbers, then $\frac{{\sin }^{3}B+{\sin }^{3}C}{\sin A\sin B\sin C}$ is equal to

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

Answer: (C)

$2$

Assume any two positive number $x$ aand $y$

So, according to the question $x,a,y$ are in A.P

So $a=\frac{x+y}{2}$ ...(1)

Also $x,b,c,y$ are in G.P

So , $b^2=xc$ and $c^2=by$ ....(2)

Therefore from (1) and (2)

$\frac{b^3+c^3}{abc}=\frac{bc(x+y)}{\frac{bc(x+y)}{2}}=2$

Now from sine rule

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k$(let)

Hence $=\frac{(\frac{\sin B}{k}{)}^3+(\frac{\sin C}{k}{)}^3}{\frac{\sin A\sin BsinC}{k^3}}=\frac{{\sin }^{3}B+{\sin }^{3}C}{\sin A\sin B\sin C}=2$