An observer in a boat finds that the angle of elevation of the top of a tower standing on the top of a cliff is $60^\circ$ and that of the top of the cliff is $30^\circ$. If the height of the tower is $60 \ m$,then the height of the cliff is ... $m$.

The horizontal distance between a tower and a building is $10 \sqrt{3}$ units. If the angle of depression of the foot of the building from the top of the tower is $60^{\circ}$ and the angle of elevation of the top of the building from the foot of the tower is $30^{\circ}$,then the sum of the heights of the tower and the building 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

The angle of elevation of the top of a tower at a point on the ground is $30^\circ$. If on walking $20 \, m$ toward the tower,the angle of elevation becomes $60^\circ$,then the height of the tower is:

$A$ spherical gas balloon of radius $16 \ m$ subtends an angle $60^{\circ}$ at the eye of the observer $A$,while the angle of elevation of its center from the eye of $A$ is $75^{\circ}$. Then the height (in $m$) of the top most point of the balloon from the level of the observer's eye is:

From the base of a pole of height $20 \text{ m}$,the angle of elevation of the top of a tower is $60^{\circ}$. The pole subtends an angle of $30^{\circ}$ at the top of the tower. Then the height of the tower is