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 :

Let a vertical tower $$A B$$ of height $$2 h$$ stands on a horizontal ground. Let from a point $$P%$$ on the ground a man can see upto height $$h$$ of the tower with an angle of elevation $$2 \alpha$$. When from $$P$$, he moves a distance $$d$$ in the direction of $$\overrightarrow{A P}$$, he can see the top $$B$$ of the tower with an angle of elevation $$\alpha$$. If $$d=\sqrt{7} h$$, then $$\tan \alpha$$ is equal to

A tower PQ stands on a horizontal ground with base $$Q$$ on the ground. The point $$R$$ divides the tower in two parts such that $$Q R=15 \mathrm{~m}$$. If from a point $$A$$ on the ground the angle of elevation of $$R$$ is $$60^{\circ}$$ and the part $$P R$$ of the tower subtends an angle of $$15^{\circ}$$ at $$A$$, then the height of the tower is :

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