$A$ vertical pole (more than $100 \, ft$ high) consists of two portions,the lower being one-third of the whole. If the upper portion subtends an angle $\tan^{-1}(\frac{1}{2})$ at a point in a horizontal plane through the foot of the pole and at a distance of $40 \, ft$ from it,then the height of the pole is......$ft$.

The angle of elevation of the top $P$ of a vertical tower $PQ$ of height $10$ from a point $A$ on the horizontal ground is $45^{\circ}$. Let $R$ be a point on $AQ$ and from a point $B$,vertically above $R$,the angle of elevation of $P$ is $60^{\circ}$. If $\angle BAQ = 30^{\circ}$,$AB = d$ and the area of the trapezium $PQRB$ is $\alpha$,then the ordered pair $(d, \alpha)$ is.

Let a vertical tower $AB$ have its end $A$ on the level ground. Let $C$ be the mid-point of $AB$ and $P$ be a point on the ground such that $AP = 2AB$. If $\angle BPC = \beta$,then $\tan \beta$ is equal to:

$A$ person observes the top of a tower from a point $A$ on the ground. The angle of elevation of the tower from this point is $60^{\circ}$. He moves $60 \ m$ in the direction perpendicular to the line joining $A$ and the base of the tower to reach point $C$. The angle of elevation of the tower from point $C$ is $45^{\circ}$. Then,the height of the tower (in metres) is

The angle of elevation of the top of a hill from a point on the horizontal plane passing through the foot of the hill is found to be $45^{\circ}$. After walking a distance of $80 \ m$ towards the top,up a slope inclined at an angle of $30^{\circ}$ to the horizontal plane,the angle of elevation of the top of the hill becomes $75^{\circ}$. Then the height of the hill (in meters) is

Two poles,$AB$ of length $a$ metres and $CD$ of length $a+b$ $(b \neq a)$ metres are erected at the same horizontal level with bases at $B$ and $D$. If $BD=x$ and $\tan \angle ACB = \frac{1}{2}$,then: