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 :

The angle of elevation of the top P of a vertical tower PQ of height 10 from a point A on the horizontal ground is $$45^{\circ}$$. Let R be a point on AQ and from a point B, vertically above $$\mathrm{R}$$, the angle of elevation of $$\mathrm{P}$$ is $$60^{\circ}$$. If $$\angle \mathrm{BAQ}=30^{\circ}, \mathrm{AB}=\mathrm{d}$$ and the area of the trapezium $$\mathrm{PQRB}$$ is $$\alpha$$, then the ordered pair $$(\mathrm{d}, \alpha)$$ is :