JEE Math Practice Question
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 :
- A.$$\frac{6}{5}$$
- B.$$\frac{9}{5}$$
- C.$$\frac{4}{3}$$
- D.$$\frac{7}{3}$$
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