$A$ tower stands vertically on the ground. The angle of elevation of the top of the tower from a point $100 \, m$ away from the base of the tower is $60^\circ$. The height of the tower is $\ldots \ldots \ldots \ldots \, m$.

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 $\frac{a \tan \theta}{1 - \tan^2 \theta}$.

In right-angled $\Delta ABC$,$m\angle B = 90^\circ$. If $AC = 20$ and $m\angle C = 30^\circ$,then $BC = \ldots$

Watching from a point $A$ on the ground $x \ m$ away from the base of the building. The height of the building is $y$ and the angle of elevation of the top of the building is found to be $25^\circ$. Then,$\ldots \ldots \ldots \ldots .$

As observed from the top of a building,the angle of depression of a man standing $10 \, m$ away from the building is found to be $45^{\circ}$. Find the height of the building (in $m$).

$A$ $1.2 \, m$ tall girl spots a balloon moving with the wind in a horizontal line at a height of $88.2 \, m$ from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is $60^{\circ}$. After some time,the angle of elevation reduces to $30^{\circ}$. Find the distance travelled by the balloon during this interval.