The angle of elevation of the top of a tower at a point on the ground is $30^\circ$. If on walking $20 \, m$ toward the tower,the angle of elevation becomes $60^\circ$,then the height of the tower is:

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.

The shadow of a tower standing on a level ground is found to be $60 \ m$ longer when the Sun's altitude is $30^{\circ}$ than when it is $45^{\circ}$. The height of the tower is

$A$ ladder rests against a wall making an angle $\alpha$ with the horizontal. The foot of the ladder is pulled away from the wall through a distance $x$,so that it slides a distance $y$ down the wall making an angle $\beta$ with the horizontal. The correct relation is

$A$ man whose eye level is $1.5 \ m$ above the ground observes the angle of elevation of a tower to be $60^\circ$. If the distance of the man from the tower is $10 \ m$,the height of the tower is:

$A$ flag-staff of $5 \, m$ high stands on a building of $25 \, m$ high. An observer is at a height of $30 \, m$. The flag-staff and the building subtend equal angles at the observer. The distance of the observer from the top of the flag-staff is