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 :
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 :
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 feet of these two poles. Let $${\pi \over 8}$$ 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 tan2$$\theta$$ is equal to
at A and B respectively on a horizontal ground.
If AD = 8 m, BC = 11 m and AB = 10 m; then the distance
(in meters) of a point M on AB from the point A such
that MD2 + MC2 is minimum is ______.