The angles of elevation of the top of a tower $A$ from the top $B$ and bottom $D$ of a building of height $a$ are $30^\circ$ and $45^\circ$ respectively. If the tower and the building stand at the same level,then the height of the tower is

$A$ tower subtends angles $\alpha, 2 \alpha$ and $3 \alpha$ respectively at points $A, B$ and $C$, all lying on a horizontal line through the foot of the tower. Then $\frac{A B}{B C}$ is equal to:

An aeroplane flying horizontally $1 \ km$ above the ground is observed at an elevation of $60^\circ$ and after $10$ seconds the elevation is observed to be $30^\circ$. The uniform speed of the aeroplane in $km/h$ is

The angular elevation of a tower $CD$ at a point $A$ due south of it is $60^\circ$ and at a point $B$ due west of $A$,the elevation is $30^\circ$. If $AB = 3 \, km$,the height of the tower is

The angle of elevation of an object from a point $P$ on the level ground is $\alpha$. Moving $d$ metres on the ground towards the object, the angle of elevation is found to be $\beta$. Then the height (in metres) of the object 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: