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.

$A$ sphere with centre $O$ sits on the top of a pole as shown in the figure. An observer on the ground is at a distance $50 \ m$ from the foot of the pole. She notes that the angles of elevation from the observer to points $P$ and $Q$ on the sphere are $30^{\circ}$ and $60^{\circ}$,respectively. Then,the radius of the sphere in metres is

$ABC$ is a triangular park with $AB = AC = 100 \text{ metres}$. $A$ vertical tower of height $h$ is situated at the mid-point $P$ of $BC$. If the angles of elevation of the top of the tower $Q$ at $A$ and $B$ are $\cot^{-1}(3\sqrt{2})$ and $\csc^{-1}(2\sqrt{2})$ respectively,then the height of the tower (in metres) 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 pillar at a point $A$ on the ground is $15^\circ$. On walking $40 \ m$ towards the pillar,the angle becomes $30^\circ$. The height of the pillar is $... \ m$.

$A$ person observes the top of a tower from a point $A$ on the ground. The angle of elevation of the tower from this point is $60^{\circ}$. He moves $60 \ m$ in the direction perpendicular to the line joining $A$ and the base of the tower to reach point $C$. The angle of elevation of the tower from point $C$ is $45^{\circ}$. Then, the height of the tower (in metres) is