$A$ person standing at a distance of $10 \ m$ from a pole observes that the angle subtended by the lower $\frac{1}{3}$rd part and the upper $\frac{2}{3}$rd part of the pole are the same. Find the height of the pole.

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:

$A$ 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:

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:

The angle of elevation of the top of a vertical tower from a point $A$,due east of it,is $45^\circ$. The angle of elevation of the top of the same tower from a point $B$,due south of $A$,is $30^\circ$. If the distance between $A$ and $B$ is $54\sqrt{2} \, \text{m}$,then the height of the tower (in metres) is:

The sum of angles of elevation of the top of a tower from two points distant $a$ and $b$ from the base and in the same straight line with it is $90^{\circ}$. Then, the height of the tower is