Trying to find if this trigonometry problem is wrong or not

Question

I know That we have a triangle ABC with $$A=90° $$ $$ b+c=18 $$ $$ \cot(B)=4\sin^2(C) $$ $$ 0<B,C<90 $$

But , upon calculating i end up a = 18, b = 18, c = 0 which should not be possible

Answer 1

$$\frac{c}{b}=\cot(B)=4\sin^2(C)=\frac{4c^2}{a^2}$$

$$a^2=4bc=b^2+c^2$$

Now use $c=18-b$ and get:

$$b^2-18b+54=0$$

$$b=9\pm3\sqrt{3}$$

Therefore, $a=6\sqrt{6}$; $b,c=9\pm3\sqrt{3}$.