The angle of elevation of the top $$\mathrm{P}$$ of a tower from the feet of one person standing due South of the tower is $$45^{\circ}$$ and from the feet of another person standing due west of the tower is $$30^{\circ}$$. If the height of the tower is 5 meters, then the distance (in meters) between the two persons is equal to

From the top $$\mathrm{A}$$ of a vertical wall $$\mathrm{AB}$$ of height $$30 \mathrm{~m}$$, the angles of depression of the top $$\mathrm{P}$$ and bottom $$\mathrm{Q}$$ of a vertical tower $$\mathrm{PQ}$$ are $$15^{\circ}$$ and $$60^{\circ}$$ respectively, $$\mathrm{B}$$ and $$\mathrm{Q}$$ are on the same horizontal level. If $$\mathrm{C}$$ is a point on $$\mathrm{AB}$$ such that $$\mathrm{CB}=\mathrm{PQ}$$, then the area (in $$\mathrm{m}^{2}$$ ) of the quadrilateral $$\mathrm{BCPQ}$$ 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 :

A horizontal park is in the shape of a triangle $$\mathrm{OAB}$$ with $$\mathrm{AB}=16$$. A vertical lamp post $$\mathrm{OP}$$ is erected at the point $$\mathrm{O}$$ such that $$\angle \mathrm{PAO}=\angle \mathrm{PBO}=15^{\circ}$$ and $$\angle \mathrm{PCO}=45^{\circ}$$, where $$\mathrm{C}$$ is the midpoint of $$\mathrm{AB}$$. Then $$(\mathrm{OP})^{2}$$ is equal to :