The angle of elevation of the top of a tower from a point A due...

Given,

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 $\cos^{- 1}\left(\dfrac{3}{\sqrt{13}}\right)$. If the distance of the point $B$ from the tower is $15$ units, then $cot\alpha$ is equal to

Now making the diagram of given value we have,

Given $OB = 15$

$cos\beta = \dfrac{3}{\sqrt{13}}$

So, $tan\beta = \dfrac{2}{3}$

Now using the value of $tan\beta$ in $△ OPB$ we get,

$⇒ tan\beta = \dfrac{h}{15}$

$⇒ \dfrac{2}{3} = \dfrac{h}{15}$, so $h = 10$

Now in $△ OAB$

$O{A}^{2} + A{B}^{2} = 225$

$⇒ O{A}^{2} + 81 = 225$

$⇒ OA = 12$

Now in triangle $△ OAP$

$tan\alpha = \dfrac{10}{12}$

$⇒ cot\alpha = \dfrac{12}{10} = \dfrac{6}{5}$