Help Pls

By polar coordinates, AB = sqrt(3^2 + 3*9 + 9^2) = 3*sqrt(13).

Let $A = (3, \theta_1)$ and $B = (9, \theta_2)$ in polar coordinates. If  $\theta_1 - \theta_2 = \dfrac{\pi}{2}$, then  find the distance $AB$

Formula:
We have two points $(r_1,\ \theta_1)$ and $( r_2,\ \theta_2)$.
The distance between two polar coordinates $\mathbf{d =\sqrt{r_1^2+r_2^2-2r_1r_2\cos(\theta_2 - \theta_1})}$

$\begin{array}{|rcll|} \hline d &=& \sqrt{r_1^2+r_2^2-2r_1r_2\cos(\theta_2 - \theta_1) } \\ d &=& \sqrt{r_1^2+r_2^2-2r_1r_2\cos\Big(-(\theta_1 - \theta_2)\Big) } \\ d &=& \sqrt{r_1^2+r_2^2+2r_1r_2\cos(\theta_1 - \theta_2) } \\ d &=& \sqrt{3^2+9^2+2*3*9\cos\left( \dfrac{\pi}{2} \right) } \\ d &=& \sqrt{9+81+54\cos\left( \dfrac{\pi}{2} \right) } \quad | \quad \cos\left( \dfrac{\pi}{2} \right) = 0 \\ d &=& \sqrt{9+81+0 } \\ d &=& \sqrt{90} \\ d &=& \sqrt{9*10} \\ d &=& \sqrt{3^2*10} \\ \mathbf{d} &=& \mathbf{3\sqrt{10}} \\ \hline \end{array}$