Practice Quantitative Aptitude questions on Height and Distance, covering trigonometric ratios, angles of elevation and depression, heights of buildings and towers, distances between objects, and real-life problems involving right-angled triangles. This test contains [question count] questions and takes approximately [duration] minutes to complete, making it useful for placement tests, TCS NQT, SSC, banking and MBA entrance preparation.
Question 1
There is a tree between houses of A and B. If the tree leans on A’s House, the tree top rests on his window which is 12 m from ground. If the tree leans on B’s House, the tree top rests on his window which is 9 m from ground. If the height of the tree is 15 m, what is distance between A’s and B’s house?
Explanation
When the tree rests on A's window, the horizontal distance = the square root of (15² - 12²) = 9 m.
When it rests on B's window, the horizontal distance = the square root of (15² - 9²) = 12 m.
The distance between the houses = 9 + 12 = 21 m.
When it rests on B's window, the horizontal distance = the square root of (15² - 9²) = 12 m.
The distance between the houses = 9 + 12 = 21 m.
Question 2
Ramesh and Suresh’s mud forts have heights 8cm and 15 cm. They are 24 cm apart. How far are the fort tops from each other?
Explanation
The two forts form the two vertical sides, and their difference in height = 15 - 8 = 7 cm, with a horizontal distance of 24 cm.
The distance between the tops = the square root of (24² + 7²) = the square root of 625 = 25 cm.
The distance between the tops = the square root of (24² + 7²) = the square root of 625 = 25 cm.
Question 3
A tree breaks and falls to the ground such that its upper part is still partially attached to its stem. At what height did it break, if the original height of the tree was 24 cm and it makes an angle of 30° with the ground?
Explanation
Let the tree break at a height h. The broken part has length 24 - h and makes an angle of 30° with the ground, so sin 30° = h24 - h.
12 = h24 - h, so 24 - h = 2h and h = 8 cm.
12 = h24 - h, so 24 - h = 2h and h = 8 cm.
Question 4
The angle of elevation of top of a tree on the bank of a river from its other bank is 60° and from a point 20 m further away from this is 30°. The width of the river is :
Explanation
Let the width of the river be w and the height of the tree be h.
tan 60° = hw, so h = w√3. tan 30° = h(w + 20), so h = (w + 20) ÷ √3.
w × √3 = (w + 20) ÷ √3, so 3w = w + 20 and w = 10 m.
tan 60° = hw, so h = w√3. tan 30° = h(w + 20), so h = (w + 20) ÷ √3.
w × √3 = (w + 20) ÷ √3, so 3w = w + 20 and w = 10 m.
Question 5
From the top of a house 32 metre high, the angle of elevation of the top of a tower is 45° and the angle of depression of the foot of the tower is 30°. The distance of the tower from the house is :
Explanation
From the top of the house (32 m high), the angle of depression of the foot of the tower is 30°, so tan 30° = 32d, which gives d = 32 × √3 = 32√3 m.
So the distance of the tower from the house is 32√3 m.
So the distance of the tower from the house is 32√3 m.
Question 6
A flag pole, which is 36.2 ft high, casts a shadow of 5 ft at 11 a.m. The shadow cast by a tower at the same time is 35 ft long. The height of the tower is:
Explanation
By similar triangles (same sun angle): tower height = 35 × (36.25) = 253.4 ft.
Question 7
A tower is 15 m high. It makes an angle of depression of 60 degrees with the bottom of an electric pole and an angle of depression of 30 degrees with the top of the pole. Calculate the height of the electric pole.
Explanation
Distance d: tan60=15/d gives d=5√3. Then 15-h=d x tan30=5√3 × (1/√3)=5, so h=10 m.
Question 8
The angle of elevation of the top of a building from the foot of a tower is 60° and the angle of elevation of the top of the tower from the foot of the building is 45°. If the tower is 150m high, find the height of the building.
Explanation
From the tower-top-elevation-45 condition, distance between them = 150/tan45 = 150 m. Building height = distance x tan60 = 150√3 = 259.80 m.
Question 9
A kite is flying at a height of 30 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 30°. Find the length of the string, assuming that there is no slack in the string.
Explanation
Length of string = height/sin30 = 300.5 = 60 m.
Question 10
An airforce officer at a signal station observes a UFO (Unidentified Flying Object) at an altitude of 1200 m flying towards his station at an angle of elevation of 30°. After exactly 60 seconds he observes the angle of elevation has changed to 60°. Find the distance covered by the plane in 1 minute.
Explanation
Distance at 30° = 1200/tan30 = 1200√3 = 2078.46 m. Distance at 60° = 1200/tan60 = 1200/√3 = 692.82 m. Distance covered = 2078.46 - 692.82 = 1385.64 m.
Question 11
Two girls on the same side of a tall building notice the angle of elevation to the top of the building to be 45° and 60° respectively. If the height of the building is known to be 200m, find the distance between the two girls.
Explanation
Farther distance = 200/tan45 = 200 m. Nearer distance = 200/tan60 = 200/√3 = 115.47 m. Difference = 200 - 115.47 = 84.53 m.
Question 12
A man is walking along a highway. He notices the top of a building subtending an angle of 45° with the ground at the point where he is standing. If the height of the building is 60 m, what is the distance in meters of the man from the building?
Explanation
Distance = height/tan45 = 601 = 60 m.
Question 13
A man standing at a point X is watching the top of a tree, which makes an angle of elevation of 45° with the man's eye. The man walks some distance towards the tree to watch the top and the angle of elevation becomes 60°. What is the distance between the base of the tree and the point?
Explanation
Without the tree's height or the distance walked given numerically, only ratios can be established between the two positions -- an absolute distance cannot be computed, so the data is inadequate.
Question 14
The angle of elevation of the sun, when the height of a tree is √3 times the length of the shadow of the tree, is
Explanation
Angle = arctan(height/shadow) = arctan(√3) = 60 degree.
Question 15
The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 11 m away from the wall. The length of the ladder is
Explanation
Length = distance/cos60 = 110.5 = 22 m.
Question 16
Two towers face each other separated by a distance d = 15 m. As seen from the top of the first tower, the angle of depression of the second tower's base is 60° and that of the top is 30°. What is the height (in meters) of the second tower?
Explanation
First tower height = 15 x tan60 = 15√3 = 25.98 m. Difference in height = 15 x tan30 = 15/√3 = 8.66 m. Second tower height = 25.98 - 8.66 = 17.32 m, which does not exactly match any listed option, so the answer is None.
Question 17
A man is walking along a straight road. He notices the top of a tower subtending an angle A = 60° with the ground at the point where he is standing. If the height of the tower is h = 20 m, then what is the distance (in meters) of the man from the tower?
Explanation
Distance = height/tan60 = 20/√3 = 11.55 m.
Question 18
A little boy is flying a kite. The string of the kite makes an angle of 30° with the ground. If the height of the kite is h = 24 m, find the length (in meters) of the string that the boy has used.
Explanation
Length of string = height/sin30 = 240.5 = 48 m.
Question 19
A ladder leaning against a wall makes an angle of 60° with the ground. If the length of the ladder is 19 m, find the distance of the foot of the ladder from the wall.
Explanation
Distance = ladder x cos60 = 19 × 0.5 = 9.5 m.
Question 20
On the same side of a tower, two objects are located. Observed from the top of the tower, their angles of depression are 30° and 60°. If the height of the tower is 100 m, the distance between the objects is:
Explanation
Distance to nearer object = 100/tan60 = 100/√3 = 57.74 m. Distance to farther object = 100/tan30 = 100√3 = 173.21 m. Difference = 173.21 - 57.74 = 115.47 m = 200/√3 m.
Question 21
Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse as observed from the two ships are 15° and 60° respectively. If the lighthouse is 150m high, the distance between the two ships is:
Explanation
Distance from first ship = 150/tan15 = 559.85 m. Distance from second ship = 150/tan60 = 86.60 m. Total distance = 646.45 m, which does not match any listed option, so the answer is None.
Question 22
From a point P on level ground, the angle of elevation of the top of a tower is 45°. If the tower is 160 m high, the distance of point P from the foot of the tower is:
Explanation
Distance = height/tan45 = 1601 = 160 m.
Question 23
The angle of elevation of the sun, when the length of the shadow of a tree is equal to the height of the tree, is:
Explanation
Angle = arctan(height/shadow) = arctan(1) = 45°.
Question 24
A man standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is 60°. When he retreats 40 m from the bank, he finds the angle to be 30°. Find the breadth of the river.
Explanation
Let breadth = b and tree height = h. tan60 = h/b and tan30 = h/(b+40). From these, b√3 = (b+40)/√3, giving 3b = b+40, so b = 20 m.
Question 25
The angle of elevation of the top of a tower at a point on the ground is 30°. On walking 12 m towards the tower, the angle of elevation becomes 60°. Find the height of the tower.
Explanation
Let height = h. Initial distance = h√3, final distance = h/√3. Their difference is 12m: h(√3 - 1/√3) = 12, giving h = 12√3/2 = 6√3 = 10.39 m.
