As observed from the top of a $510 \ m$ high tower,the angles of depression of two houses situated in the east and west of the tower are found to be $30^{\circ}$ and $60^{\circ}$ respectively. Find the distance between the two houses (in $m$).

An observer $1.5\, m$ tall is $28.5\, m$ away from a tower. The angle of elevation of the top of the tower from his eyes is $45^\circ$. Then the height of the tower is ............ $m$.

In $\Delta ABC$,$\angle B$ is a right angle. If $AC = 20$ and $BC = 10$,then $m\angle ACB = \ldots$ (in $^{\circ}$)

$A$ person is watching from a point $A$ on the ground,which is $x \ m$ away from the base of a building. The height of the building is $y \ m$ and the angle of elevation of the top of the building is found to be $70^\circ$. Then:

Watching from the top of a tower,the angles of depression of two houses $A$ and $B$ are found to be $25^{\circ}$ and $40^{\circ}$ respectively. Then,.......

$A$ window of a house is $h$ meters above the ground. From the window,the angles of elevation and depression of the top and the bottom of another house situated on the opposite side of the lane are found to be $\alpha$ and $\beta$,respectively. Prove that the height of the other house is $h(1 + \tan \alpha \cot \beta)$ meters.