The angle of elevation of the top of a pillar at a point $A$ on the ground is $15^\circ$. On walking $40 \ m$ towards the pillar,the angle becomes $30^\circ$. The height of the pillar is $... \ m$.

The angle of elevation of an object on a hill is observed from a certain point in the horizontal plane through its base to be $30^{\circ}$. After walking $120 \ m$ towards it on level ground, the angle of elevation is found to be $60^{\circ}$. Then the height of the object (in metres) is:

The angle of elevation of a stationary cloud from a point $2500 \ m$ above a lake is $15^{\circ}$ and from the same point the angle of depression of its reflection in the lake is $45^{\circ}$. The height (in metres) of the cloud above the lake, given that $\cot 15^{\circ}=2+\sqrt{3}$, is

$A$ horizontal park is in the shape of a triangle $OAB$ with $AB = 16$. $A$ vertical lamp post $OP$ is erected at the point $O$ such that $\angle PAO = \angle PBO = 15^{\circ}$ and $\angle PCO = 45^{\circ}$,where $C$ is the midpoint of $AB$. Then $(OP)^{2}$ 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 top of a hill observed from the top and bottom of a building of height $h$ is at the angle of elevation $p$ and $q$ respectively. The height of the hill is