$A$ tower subtends an angle of $30^\circ$ at a point distant $d$ from the foot of the tower and on the same level as the foot of the tower. At a second point $h$ vertically above the first,the angle of depression of the foot of the tower is $60^\circ$. The height of the tower is:

$A$ 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

$A$ tower of height $b$ subtends an angle $\alpha$ at a point $O$ on the level of the foot of the tower and at a distance $a$ from the foot of the tower. If a pole of height $p$ mounted on the tower also subtends an equal angle $\alpha$ at $O$,the height of the pole $p$ is:

$A$ man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizontal road. If it takes $18 \text{ min}$ for the angle of depression of the car to change from $30^\circ$ to $45^\circ$,then after this,the time taken (in min) by the car to reach the foot of the tower is:

The angle of elevation of a tower from point $P$ due north of it is $30^\circ$ and from a point $Q$ due east of point $P$ is $\theta$. If the distance of the foot of the tower from point $P$ is $3 \ m$ and $PQ = 3 \ m$,then $\theta$ is equal to-

$A$ 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......$ft$.