Two poles standing on a horizontal ground are of heights $5 \, m$ and $10 \, m$ respectively. The line joining their tops makes an angle of $15^o$ with the ground. Then the distance (in $m$) between the poles is:

$A$ tower $PQ$ stands on a horizontal ground with base $Q$ on the ground. The point $R$ divides the tower into two parts such that $QR = 15 \, m$. If from a point $A$ on the ground,the angle of elevation of $R$ is $60^{\circ}$ and the part $PR$ of the tower subtends an angle of $15^{\circ}$ at $A$,then the height of the tower is:

From the top of a hill $h$ metres high, the angles of depression of the top and the bottom of a pillar are $\alpha$ and $\beta$ respectively. The height (in metres) of the pillar is

Two vertical poles of heights $20 \ m$ and $80 \ m$ stand apart on a horizontal plane. The height (in meters) of the point of intersection of the lines joining the top of each pole to the foot of the other,from this horizontal plane is:

$A$ 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

$A$ 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: