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:

The length of the shadow of a pole inclined at $10^\circ$ to the vertical towards the sun is $2.05 \text{ m}$,when the elevation of the sun is $38^\circ$. The length of the pole is

The angle of elevation of an object from a point $P$ on the level ground is $\alpha$. Moving $d$ metres on the ground towards the object,the angle of elevation is found to be $\beta$. Then the height (in metres) of the object is

Let $AB$ and $PQ$ be two vertical poles,$160 \ m$ apart from each other. Let $C$ be the middle point of $B$ and $Q$,which are the feet of these two poles. Let $\frac{\pi}{8}$ and $\theta$ be the angles of elevation from $C$ to $P$ and $A$,respectively. If the height of pole $PQ$ is twice the height of pole $AB$,then $\tan^{2} \theta$ is equal to

$A$ horizontal park is in the shape of a triangle $OAB$ with $AB = 16$. $A$ vertical lamp post $OP$ is erected at the point $O$ such that $\angle PAO = \angle PBO = 15^{\circ}$ and $\angle PCO = 45^{\circ}$,where $C$ is the midpoint of $AB$. Then $(OP)^{2}$ is equal to.

$ABCD$ is a rectangular field. $A$ vertical lamp post of height $12 \ m$ stands at the corner $A$. If the angle of elevation of its top from $B$ is $60^{\circ}$ and from $C$ is $45^{\circ}$,then the area of the field is