If the angle of elevation of the top of a tower at a distance of $500 \, m$ from its foot is $30^\circ$,then the height of the tower is:

The angle of elevation of a point situated at a distance of $70 \ m$ from the base of a tower is $45^\circ$. The height of the tower is ..... $m$.

Two vertical poles are $150 \ m$ apart and the height of one is three times that of the other. If from the middle point of the line joining their feet,an observer finds the angles of elevation of their tops to be complementary,then the height of the shorter pole (in meters) is

$A$ tower subtends an angle $\alpha$ at a point $A$ in the plane of its base,and the angle of depression of the foot of the tower from a point $P$ at a height $l$ meters vertically above $A$ is $\beta$. The height of the tower is

$A$ pole stands vertically inside a triangular park $\Delta ABC$. Let the angle of elevation of the top of the pole from each corner of the park be $\frac{\pi}{3}$. If the radius of the circumcircle of $\Delta ABC$ is $2$,then the height of the pole is equal to:

$A$ tower $T_1$ of height $60 \, m$ is located exactly opposite to a tower $T_2$ of height $80 \, m$ on a straight road. From the top of $T_1$,if the angle of depression of the foot of $T_2$ is twice the angle of elevation of the top of $T_2$,then the width (in $m$) of the road between the feet of the towers $T_1$ and $T_2$ is