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:

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

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

$A$ tree is broken by wind,its upper part touches the ground at a point $10 \ m$ from the foot of the tree and makes an angle of $45^\circ$ with the ground. The total length of the tree is:

The angle of elevation of the top of a tower from a point $A$ due north of it is $\alpha$ and from a point $B$ at a distance of $9$ units due west of $A$ is $\cos^{-1}\left(\frac{3}{\sqrt{13}}\right)$. If the distance of the point $B$ from the tower is $15$ units,then $\cot \alpha$ is equal to.

In the figure,$\theta_1+\theta_2=\frac{\pi}{2}$ and $\sqrt{3}(BE)=4(AB)$. If the area of $\triangle CAB$ is $2\sqrt{3}-3 \text{ unit}^2$,when $\frac{\theta_2}{\theta_1}$ is the largest,then the perimeter (in unit) of $\triangle CED$ is equal to $...........$.