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:

An aeroplane is flying at a constant speed,parallel to the horizontal ground at a height of $5 \text{ km}$. $A$ person on the ground observed that the angle of elevation of the plane is changed from $15^{\circ}$ to $30^{\circ}$ in the duration of $50 \text{ seconds}$. The speed of the plane (in $\text{kmph}$) is:

Let $10$ vertical poles standing at equal distances on a straight line,subtend the same angle of elevation $\alpha$ at a point $O$ on this line and all the poles are on the same side of $O$. If the height of the longest pole is $h$ and the distance of the foot of the smallest pole from $O$ is $a$,then the distance between two consecutive poles is:

The angle of elevation of the top of a tower at a point on the ground is $30^\circ$. If on walking $20 \, m$ toward the tower,the angle of elevation becomes $60^\circ$,then the height of the tower is:

In a triangular park $PQR$,$PQ = PR = 200 \ m$. $A$ $TV$ tower stands at the mid-point $M$ of $QR$. If the angles of elevation of the top of the tower at $P, Q,$ and $R$ are $45^o, 30^o,$ and $30^o$ respectively,then the height of the tower (in $m$) is:

$A$ man is observing,from the top of a tower,a boat speeding towards the tower from a certain point $A$,with uniform speed. At that point,the angle of depression of the boat from the man's eye is $30^{\circ}$ (ignore the man's height). After sailing for $20 \text{ seconds}$ towards the base of the tower (which is at the level of water),the boat reaches a point $B$,where the angle of depression is $45^{\circ}$. Then,the time taken (in seconds) by the boat from $B$ to reach the base of the tower is: