JEE Main 2022 — Heights and Distances Question with Solution

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 ${\cos }^{-1}\left(\frac{3}{\sqrt{13}}\right)$. 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 $OB=15$

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

So, $\tan \beta =\frac{2}{3}$

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

$\Rightarrow \tan \beta =\frac{h}{15}$

$\Rightarrow \frac{2}{3}=\frac{h}{15}$, so $h=10$

Now in $△OAB$

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

$\Rightarrow O{A}^2+81=225$

$\Rightarrow OA=12$

Now in triangle $△OAP$

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

$\Rightarrow \cot \alpha =\frac{12}{10}=\frac{6}{5}$