Two pillars of equal height stand on either side of a roadway which is $60 \, m$ wide. At a point in the roadway between the pillars,the angles of elevation of the top of the pillars are $60^\circ$ and $30^\circ$. The height of the pillars is

$A$ man is observing,from the top of a tower,a boat speeding towards the tower from a certain point $A$,with uniform speed. At that point,the angle of depression of the boat from the man's eye is $30^{\circ}$ (ignore the man's height). After sailing for $20 \text{ seconds}$ towards the base of the tower (which is at the level of water),the boat reaches a point $B$,where the angle of depression is $45^{\circ}$. Then,the time taken (in seconds) by the boat from $B$ to reach the base of the tower is:

An observer in a boat finds that the angle of elevation of the top of a tower standing on the top of a cliff is $60^\circ$ and that of the top of the cliff is $30^\circ$. If the height of the tower is $60 \ m$,then the height of the cliff is ... $m$.

Let a vertical tower $AB$ of height $2h$ stand on a horizontal ground. From a point $P$ on the ground,a man can see up to height $h$ of the tower with an angle of elevation $2\alpha$. When he moves a distance $d$ from $P$ in the direction of $\overline{AP}$,he can see the top $B$ of the tower with an angle of elevation $\alpha$. If $d=\sqrt{7}h$,then $\tan \alpha$ is equal to:

$ABC$ is a triangular park with $AB = AC = 100 \text{ metres}$. $A$ vertical tower of height $h$ is situated at the mid-point $P$ of $BC$. If the angles of elevation of the top of the tower $Q$ at $A$ and $B$ are $\cot^{-1}(3\sqrt{2})$ and $\csc^{-1}(2\sqrt{2})$ respectively,then the height of the tower (in metres) is

The upper part of a tree broken by the wind makes an angle of $30^\circ$ with the ground. The distance from the root to the point where the top of the tree touches the ground is $10 \, m$. What was the original height of the tree in meters?