If triangle ABC satisfies the condition $2\sin{A} + \sin{B} = 2\sin{C}$, find the minimum value of $\frac{5}{\sin{A}}+\frac{9}{\sin{C}}$.

Question

I got $2\sin\frac{A+C}{2}\cos\frac{A+C}{2}=4\sin\frac{C-A}{2}\cos\frac{A+C}{2}$

then I got $\sin\frac{A+C}{2}=2\sin\frac{C-A}{2}$

so I can get $3\tan\frac{A}{2}=\tan\frac{C}{2}$

let $t = \tan\frac{A}{2}$, so $\sin{A}=\frac{2t}{t^2+1}$ , $\sin{C}=\frac{6t}{9t^2+1}$

So I can get the minimum is 16

But I think this method is too complicated. Is there a simpler way?