Height and Distance Questions

Height and Distance Questions are a helpful topic for students who want to learn trigonometry in an easy way. Questions on Height and Distance help us to find the unknown heights and distances using angles and simple formulas. The topic is very important as it builds a strong base in maths and improves the problem solving skills. Students can practice various questions to understand how the concept works in real examples.

Height and Distance Questions with Answers

Q1: A ladder leans against a wall making 60° with the ground. The foot of the ladder is 5 m from the wall. Find the height.

tan 60° = h/5

√3 = h/5

h = 5√3 = 8.66 m

Answer: Height = 5√3 m ≈ 8.66 m

Q2: The shadow of a pole is 10 m long when the sun's elevation is 30°. Find the height of the pole.

tan 30° = h/10

1/√3 = h/10

h = 10/√3 = 10√3/3 ≈ 5.77 m

Answer: Height = 10/√3 m ≈ 5.77 m

Q3: From a point 20 m away from the base of a tower, the angle of elevation of the top is 45°. Find the height.

tan 45° = h/20

1 = h/20

h = 20 m

Answer: Height = 20 m

Note: When angle = 45°, height = distance always.

Q4: A man standing on top of a 60 m high cliff observes two ships at angles of depression of 30° and 45°. Find the distance between the ships.

Solution:

Man at M (top of cliff, height = 60 m)

For Ship S2 (angle 45°):

tan 45° = 60/d2

d2 = 60 m

For Ship S1 (angle 30°):

tan 30° = 60/d1

1/√3 = 60/d1

d1 = 60√3 m

Distance between ships = d1 - d2

= 60√3 - 60

= 60(√3 - 1)

= 60(1.732 - 1)

= 60 × 0.732

= 43.92 m

Answer: Distance between ships = 60(√3 - 1) m ≈ 43.92 m

Q5: Two poles of equal heights stand on either side of a road 80 m wide. From a point between them on the road, the angles of elevation of the tops are 60° and 30°. Find the heights and position of the point.

Let AP = x, PB = 80 - x

From point P:

tan 60° = h/x → √3 = h/x → h = √3x  ...(1)

tan 30° = h/(80-x) → 1/√3 = h/(80-x) → h = (80-x)/√3  ...(2)

From (1) and (2):

√3x = (80-x)/√3

3x = 80 - x

4x = 80

x = 20 m

h = √3 × 20 = 20√3 m

Answer: Height = 20√3 m ≈ 34.64 m

Point is 20 m from the first pole.

Q6: A vertical tower stands on a horizontal plane. From a point 150 m from the base, the angle of elevation of the top is 30°. After walking some distance towards the tower, the angle becomes 60°. Find the distance walked.

Let initial point A be 150 m from base B.

Let C be the point after walking.

Let CB = d.

From A: tan 30° = h/150

h = 150/√3 = 50√3  ...(1)

From C: tan 60° = h/d

h = d√3  ...(2)

From (1) and (2):

50√3 = d√3

d = 50 m

Distance walked = AC = 150 - 50 = 100 m

Answer: Distance walked = 100 m

Solved Height and Distance Questions

Q7: The angle of elevation of the top of a tree from a point on the ground 40 m from the tree is 30°. Find the height.

h = 40 × tan 30°

h = 40 × (1/√3)

h = 40/√3 = 40√3/3 ≈ 23.09 m

Answer: ≈ 23.09 m

Q8: From the top of a lighthouse 75 m high, the angle of depression of a boat is 30°. Find the distance of the boat from the lighthouse.

tan 30° = 75/d

1/√3 = 75/d

d = 75√3 m ≈ 129.9 m

Answer: 75√3 m ≈ 129.9 m

Q9: A 1.5 m tall person stands 28.5 m from a building. The angle of elevation of the top of the building from the top of his head is 45°. Find the height of the building.

   

Height above person's head = 28.5 × tan 45°

= 28.5 × 1 = 28.5 m

Total building height = 28.5 + 1.5 = 30 m

Answer: Height = 30 m

Q10: From a window 20 m high above the ground, the angle of elevation of the top of a building across the road is 60°, and the angle of depression of its base is 30°. Find the height of the building.

Let d = horizontal distance between buildings.

Angle of depression to base = 30°:

tan 30° = 20/d

d = 20√3 m

Angle of elevation to top = 60°:

Let extra height above window = h1

tan 60° = h1/d

h1 = d × √3 = 20√3 × √3 = 60 m

Total height of building = 20 + 60 = 80 m

Answer: Height of building = 80 m

Q11: Two men on opposite sides of a tower of height 60 m observe the top at angles 30° and 60°. Find the distance between the two men.

Man 1 (angle 30°):

tan 30° = 60/d1

d1 = 60√3 m

Man 2 (angle 60°):

tan 60° = 60/d2

d2 = 60/√3 = 20√3 m

Total distance = d1 + d2

= 60√3 + 20√3

= 80√3 m

≈ 138.56 m

Answer: 80√3 m

Height and Distance MCQs

Q1: A pole is 30 m high. Angle of elevation = 45°. Distance from pole = ?

tan 45° = 30/d → d = 30 m

a) 15m  b) 30m  c) 30√3m  d) 60m

Answer: b) 30 m

Q2: If angle of elevation = 60° and distance = 10 m, height = ?

h = 10 × tan 60° = 10√3 m

a) 10m  b) 10/√3  c) 10√3  d) 5√3

Answer: c) 10√3 m 

Q3: When shadow equals height of pole, angle of elevation of sun is:

tan θ = h/h = 1 → θ = 45°

a) 30°  b) 45°  c) 60°  d) 90°

Answer: b) 45°

Q4: A 20 m tall building casts a shadow of 20√3 m. The angle of elevation of the sun is:

tan θ = 20/(20√3) = 1/√3

θ = 30°

a) 45°  b) 60°  c) 30°  d) 90°

Answer: c) 30°

Q5: The angle of depression from a 100 m cliff to a boat is 60°. Distance of boat from cliff base = ?

tan 60° = 100/d → d = 100/√3 = 100√3/3 m

a) 100m  b) 100√3m  c) 100/√3m  d) 50m

Answer: c) 100/√3 m

Q6: From point A, the angle of elevation of a tower is 30°. Moving 20 m closer, it becomes 60°. Height of tower = ?

Let distance from tower at B = d

h = d×tan60° = d√3

Also h = (d+20)×tan30° = (d+20)/√3

d√3 = (d+20)/√3

3d = d+20

2d = 20, d = 10

h = 10√3 m

a) 10m  b) 10√3m  c) 20m  d) 20√3m

Answer: b) 10√3 m

Q7: From a 7 m tall building, the angle of elevation to the top of a cable tower is 60° and angle of depression to its foot is 45°. Height of tower = ?

Horizontal distance d:

tan 45° = 7/d → d = 7 m

Height above building top:

h1 = 7 × tan 60° = 7√3 m

Total tower height = 7 + 7√3 = 7(1+√3) m

a) 7m  b) 7√3m  c) 7(1+√3)m  d) 14m

Answer: c) 7(1+√3) m

Height and Distance Word Problems

Q8: A person standing 50 m from a building sees the top at 45°. A flag at the top is seen at 60°. Find the length of the flag.

Building height = 50 × tan 45° = 50 m

Height to flag top = 50 × tan 60° = 50√3 m

Flag length = 50√3 - 50 = 50(√3 - 1) ≈ 36.6 m

Answer: 50(√3−1) m

Q9: A TV tower stands vertically on a bank of a canal. From a point on the other bank, the angle of elevation is 60°. From a point 20 m farther, it is 30°. Find the height.

tan 60° = h/d → h = d√3  ...(1)

tan 30° = h/(d+20) → h = (d+20)/√3  ...(2)

d√3 = (d+20)/√3

3d = d + 20

d = 10 m

h = 10√3 m

Answer: Height = 10√3 m ≈ 17.32 m

Q12: A circus artist climbs a 20 m long rope which is tightly stretched and tied from the top of a vertical pole to the ground. If the angle made by the rope with the ground is 30°, find the height of the pole.

sin 30° = h/20

1/2 = h/20

h = 10 m

Answer: Height of pole = 10 m

Q13: An observer 1.5 m tall is 20.5 m away from a chimney. The angle of elevation of the top of the chimney from his eyes is 45°. Find the height of the chimney.

Height above observer's eyes = 20.5 × tan 45°

= 20.5 × 1

= 20.5 m

Total chimney height = 20.5 + 1.5 = 22 m

Answer: Height = 22 m

Previous-Year Questions

Q14 (Board 2020): The angle of elevation of a cloud from a point 60 m above a lake is 30° and the angle of depression of the cloud's reflection in the lake is 60°. Find the height of the cloud above the lake surface.

Let cloud height above lake = h

Point P is 60 m above lake.

Cloud C is at height h.

Reflection R is at depth h below lake.

From P to cloud C:

tan 30° = (h-60)/d → d = (h-60)√3  ...(1)

From P to reflection R:

tan 60° = (h+60)/d → d = (h+60)/√3  ...(2)

From (1) and (2):

(h-60)√3 = (h+60)/√3

3(h-60) = h+60

3h - 180 = h + 60

2h = 240

h = 120 m

Answer: Height of cloud = 120 m above lake

Download PDF - Practice Questions on Height and Distance

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