The angle of elevation of the top of a hill from a point on the ground is $30^\circ$. After walking $30 \, m$ towards the hill,the angle of elevation becomes $45^\circ$. What is the height of the hill? (in $m$)

$A$ 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 $\ldots \ldots \ldots \ldots \, m$.

On walking for search of a ball $x$ metres on a hill making an angle of $30^{\circ}$ with the ground,one can reach a height of $y$ metres from the ground. Then $\ldots \ldots \ldots . . .$

In $\Delta ABC$,$\angle B$ is a right angle. If $AC = 16$ and $BC = 8$,then $m \angle C = \dots$ (in $^{\circ}$)

$A$ $1.2 \, m$ 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.

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