The angle of elevation of the top of a tower from a point A due north of it is $$\alpha$$ and from a point B at a distance of 9 units due west of A is $$\cos^{-1}\left(\frac{3}{\sqrt{13}}\right)$$. If the distance of the point B from the tower is 15 units, then $$\cot\alpha$$ is equal to

Let the base of the tower be at point $$T$$ and its height be $$h$$. Point $$A$$ is due north of the tower, and point $$B$$ is 9 units due west of $$A$$.

We place $$T$$ at the origin. Since $$A$$ is due north of $$T$$, let the distance $$TA = d$$. Since $$B$$ is 9 units due west of $$A$$, the distance $$TB = \sqrt{d^2 + 81}$$.

We are given that the distance from $$B$$ to the tower (i.e., $$TB$$) is 15 units. So $$\sqrt{d^2 + 81} = 15$$, giving $$d^2 + 81 = 225$$, hence $$d^2 = 144$$ and $$d = 12$$.

Now, the angle of elevation from $$B$$ is $$\beta = \cos^{-1}\!\left(\dfrac{3}{\sqrt{13}}\right)$$. We have $$\cos\beta = \dfrac{3}{\sqrt{13}}$$, so $$\sin\beta = \dfrac{2}{\sqrt{13}}$$ and $$\tan\beta = \dfrac{2}{3}$$.

Since $$\tan\beta = \dfrac{h}{TB} = \dfrac{h}{15}$$, we get $$h = 15 \times \dfrac{2}{3} = 10$$.

The angle of elevation from $$A$$ is $$\alpha$$, with $$\tan\alpha = \dfrac{h}{d} = \dfrac{10}{12} = \dfrac{5}{6}$$.

Hence $$\cot\alpha = \dfrac{1}{\tan\alpha} = \dfrac{6}{5}$$.

Hence, the correct answer is Option A.