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:

$A$ 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:

$A$ vertical pole subtends an angle $\tan ^{-1}\left(\frac{1}{2}\right)$ at a point $P$ on the ground. If the angles subtended by the upper half and the lower half of the pole at $P$ are respectively $\alpha$ and $\beta$, then $(\tan \alpha, \tan \beta)$ is equal to

$A$ man is walking towards a vertical pillar in a straight path,at a uniform speed. At a certain point $A$ on the path,he observes that the angle of elevation of the top of the pillar is $30^o$. After walking for $10 \text{ minutes}$ from $A$ in the same direction,at a point $B$,he observes that the angle of elevation of the top of the pillar is $60^o$. Then the time taken (in minutes) by him,from $B$ to reach the pillar,is:

The top of a hill observed from the top and bottom of a building of height $h$ is at the angle of elevation $p$ and $q$ respectively. The height of the hill is

An aeroplane flying at a constant speed,parallel to the horizontal ground,$\sqrt{3} \ km$ above it,is observed at an elevation of $60^\circ$ from a point on the ground. If,after five seconds,its elevation from the same point is $30^\circ$,then the speed (in $km/hr$) of the aeroplane is