In $\Delta ABC$,$m \angle B = 90^{\circ}$. If $m \angle C = \theta$,then $\tan \theta = \dots$

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 $\ldots \ldots \ldots . . .$

$A$ pole stands on the ground. Watching from the top of the pole,the angle of depression of a stone on the ground $10 \, m$ away from the pole is $60^{\circ}$. Then the height of the pole is $\ldots \ldots \ldots \, m$. (Take $\sqrt{3} = 1.73$.)

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

$A$ vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height $h$. At a point on the plane,the angles of elevation of the bottom and the top of the flag staff are $\alpha$ and $\beta$,respectively. Prove that the height of the tower is $\left(\frac{h \tan \alpha}{\tan \beta-\tan \alpha}\right)$.

$A$ ladder leans against a wall such that its lower end makes an angle of measure $\theta$ with the ground. If its lower end is $a \ m$ away from the base of the wall,then the ladder reaches the wall at a height of $\ldots \ldots \ldots \ldots \ldots \ m$.