Height and Distance

A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30^o. After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60^o. Then the time taken (in minutes) by him, from B to reach the pillar, is :

A person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank of the river is $${60^ \circ }$$ and when he retires $$40$$ meters away from the tree the angle of elevation becomes $${30^ \circ }$$. The breadth of the river is :

A tower stands at the centre of a circular park. $$A$$ and $$B$$ are two points on the boundary of the park such that $$AB(=a)$$ subtends an angle of $${60^ \circ }$$ at the foot of the tower, and the angle of elevation of the top of the tower from $$A$$ or $$B$$ is $${30^ \circ }$$. The height of the tower is :

$$AB$$ is a vertical pole with $$B$$ at the ground level and $$A$$ at the top. $$A$$ man finds that the angle of elevation of the point $$A$$ from a certain point $$C$$ on the ground is $${60^ \circ }$$. He moves away from the pole along the line $$BC$$ to a point $$D$$ such that $$CD=7$$ m. From $$D$$ the angle of elevation of the point $$A$$ is $${45^ \circ }$$. Then the height of the pole is :

$$ABCD$$ is a trapezium such that $$AB$$ and $$CD$$ are parallel and $$BC \bot CD.$$ If $$\angle ADB = \theta ,\,BC = p$$ and $$CD = q,$$ then AB is equal to:

A bird is sitting on the top of a vertical pole $$20$$ m high and its elevation from a point $$O$$ on the ground is $${45^ \circ }$$. It files off horizontally straight away from the point $$O$$. After one second, the elevation of the bird from $$O$$ is reduced to $${30^ \circ }$$. Then the speed (in m/s) of the bird is :

If the angles of elevation of the top of a tower from three collinear points $$A, B$$ and $$C,$$ on a line leading to the foot of the tower, are $${30^ \circ }$$, $${45^ \circ }$$ and $${60^ \circ }$$ respectively, then the ratio, $$AB:BC,$$ is :

Two vertical poles are 150 m apart and the height of one is three times that of the other. If from the middle point of the line joining their feet, an observer finds the angles of elevation of their tops to be complementary, then the height of the shorter pole (in meters) is :

Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = 2AB. If $$\angle$$BPC = $$\beta$$, then tan$$\beta$$ is equal to:

PQR is a triangular park with PQ = PR = 200 m. A T.V. tower stands at the mid-point of QR. If the angles of elevation of the top of the tower at P, Q and R are respectively 45$$^\circ$$, 30$$^\circ$$ and 30$$^\circ$$, then the height of the tower (in m) is :

An aeroplane flying at a constant speed, parallel to the horizontal ground, $$\sqrt 3$$ kmabove it, is obsered at an elevation of $${60^o}$$ from a point on the ground. If, after five seconds, its elevation from the same point, is $${30^o}$$, then the speed (in km / hr) of the aeroplane, is :

A tower T₁ of height 60 m is located exactly opposite to a tower T₂ of height 80 m on a straight road. Fromthe top of T₁, if the angle of depression of the foot of T₂ is twice the angle of elevation of the top of T₂, then the width (in m) of the road between the feetof the towers T₁ and T₂ is :

A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizontal road. If it takes 18 min. for the angle of depression of the car to change from 30^o to 45^o ; then after this, the time taken (in min.) by the car to reach the foot of the tower, is :

Consider a triangular plot ABC with sides AB = 7m, BC = 5m and CA = 6m. A vertical lamp-post at the mid point D of AC subtends an angle 30^o at B. The height (in m) of the lamp-post is -

The angle of elevation of the top of a vertical tower standing on a horizontal plane is observed to be 45^o from a point A on the plane. Let B be the point 30 m vertically above the point A. If the angle of elevation of the top of the tower from B be 30^o, then the distance (in m) of the foot of the tower from the point A is :

ABC is a triangular park with AB = AC = 100 metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at A and B are cot^–1 (3$$\sqrt 2$$ ) and cosec^–1 (2$$\sqrt 2$$ ) respectively, then the height of the tower (in metres) is :

Two poles standing on a horizontal ground are of heights 5m and 10 m respectively. The line joining their tops makes an angle of 15º with ground. Then the distance (in m) between the poles, is :-

Two vertical poles of heights, 20 m and 80 m stand a part on a horizontal plane. The height (in meters) of the point of intersection of the lines joining the top of each pole to the foot of the other, from this horizontal plane is :

If the angle of elevation of a cloud from a point P which is 25 m above a lake be 30^o and the angle of depression of reflection of the cloud in the lake from P be 60^o, then the height of the cloud (in meters) from the surface of the lake is :