The angle of depression of a ship from the top of a tower $30 \, m$ high is $60^\circ$. Find the distance of the ship from the base of the tower.

$A$ balloon is observed simultaneously from three points $A, B$ and $C$ on a straight road directly under it. The angular elevation at $B$ is twice and at $C$ is thrice that of $A$. If the distance between $A$ and $B$ is $200$ metres and the distance between $B$ and $C$ is $100$ metres,then the height of the balloon 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

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 $...........$.

The shadow of a tower is found to be $60 \ m$ shorter when the sun's altitude changes from $30^{\circ}$ to $60^{\circ}$. The height of the tower from the ground is approximately equal to......$m$

$A$ 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.