Observing an object from the point of observation,if one gets the angle of depression,then the object under observation is.........

Watching from the top of a building $40 \, m$ high,the angle of depression of a moving car increases from $30^{\circ}$ to $45^{\circ}$ in some interval of time. Find the distance covered by the car in that interval of time. (in $m$)

Watching from the top of an $80 \, m$ high hill,the angle of elevation of the top of a tower is $30^\circ$ and the angle of depression of the base of the tower is $45^\circ$. Find the height of the tower in $m$.

From the top of a hill $100\, m$ high,the angles of depression of the top and the bottom of a tower are observed to be $30^{\circ}$ and $45^{\circ}$ respectively. Find the height of the tower in $m$.

From the top of a tower,the angles of depression of points $A$ and $B$ in the same direction of the tower are found to be $\theta$ and $(90^\circ - \theta)$ respectively. If $B$ is nearer the tower than $A$ and $AB = a$,then prove that the height of the tower is $\frac{a \tan \theta}{1 - \tan^2 \theta}$.

When observed from the top of a tower,the angles of depression of two houses $A$ and $B$ in the Eastern and Western directions are $30^{\circ}$ and $60^{\circ}$ respectively. Then........