$P$ is a point on the segment joining the feet of two vertical poles of heights $a$ and $b$. The angles of elevation of the tops of the poles from $P$ are $45^{\circ}$ each. Then,the square of the distance between the tops of the poles is

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

$A$ 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

From a point on the level ground,the angle of elevation of the top of a pole is $30^{\circ}$. On moving $20 \ m$ nearer to the pole,the angle of elevation becomes $45^{\circ}$. The height of the pole (in metres) 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 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.