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# 16+ Heights & Distance Questions for CAT with SOLUTIONS

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## What are some CAT Heights & Distance Practice Questions?

Q1. A 30-meter tall building casts a 40-meter long shadow. At the same time, a nearby pole casts a 6-meter long shadow. What is the height of the pole?

Answer: the height of the pole is 4.5 meters.

Solution: The triangles formed by the building and its shadow, and the pole and its shadow, are similar triangles because they share the same angle with the ground (the angle of elevation of the sun). In similar triangles, the ratios of corresponding sides are equal.

Building's height / Building's shadow length = Pole's height / Pole's shadow length

30 / 40 = x / 6

Cross multiply:

30 × 6 = 40 × x

180 = 40x

Solve for x:

x = 180 ÷ 40

x = 4.5

Therefore, the height of the pole is 4.5 meters.

Q2. From the top of a lighthouse 120 meters high, the angles of depression of two ships are 30° and 45°. If both ships are on the same side of the lighthouse, find the distance between the ships.

Answer: the distance between the two ships is approximately 87.85 meters.

Solution: Let the distance between the ships be d meters.

We can calculate d using the tangent function:

tan(30°) = 120 / (d + 120 × cot(45°))

tan(30°) = 120 / (d + 120)

Solve for d:

d + 120 = 120 / tan(30°)

d + 120 = 120 × tan(60°)

d + 120 = 120 × √3

d = 120 × √3 - 120

d = 120(√3 - 1)

d ≈ 87.85

Therefore, the distance between the two ships is approximately 87.85 meters.

Q3. A kite is flying at a height of 80 meters above the ground. The string makes an angle of 60° with the ground. Find the length of the string.

Answer: the length of the kite string is approximately 92.38 meters.

Solution: Let the length of the string be L meters.

We can calculate L using the sine function:

sin(60°) = 80 / L

Solve for L:

L = 80 / sin(60°)

L = 80 / (√3 / 2)

L = 80 × (2 / √3)

L = 160 / √3

L ≈ 92.38

Therefore, the length of the kite string is approximately 92.38 meters.

Q4. The angle of elevation of the top of a tower from a point A on the ground is 30°. From another point B, 50 meters closer to the foot of the tower, the angle of elevation is 60°. Find the height of the tower.

Answer: the height of the tower is approximately 84.64 meters.

Solution: Let the height of the tower be h meters and the distance between point A and the foot of the tower be d meters.

From point A, we have:

tan(30°) = h / d

From point B, we have:

tan(60°) = h / (d - 50)

Divide the two equations:

tan(60°) / tan(30°) = (h / (d - 50)) / (h / d)

√3 = d / (d - 50)

Solve for d:

√3(d - 50) = d

√3d - 50√3 = d

√3d - d = 50√3

d(√3 - 1) = 50√3

d = 50√3 / (√3 - 1)

d = 50√3 × (√3 + 1) / (√3 - 1)(√3 + 1)

d = 50(3 + √3) / 2

d ≈ 146.65

Now, use the value of d to find h:

tan(30°) = h / d

h = d × tan(30°)

h ≈ 146.65 × (1 / √3)

h ≈ 84.64

Therefore, the height of the tower is approximately 84.64 meters.

## What are the must-do Heights & Distance questions for the CAT exam?

Q5. A ladder 10 meters long rests against a vertical wall. If the ladder makes an angle of 60° with the ground, find the distance of the foot of the ladder from the wall.

Answer: the distance of the foot of the ladder from the wall is 5 meters.

Solution: Let the distance of the foot of the ladder from the wall be d meters.

We can calculate d using the cosine function:

cos(60°) = d / 10

Solve for d:

d = 10 × cos(60°)

d = 10 × (1/2)

d = 5

Therefore, the distance of the foot of the ladder from the wall is 5 meters.

Q6. The angles of elevation of the top of a tower from two points at a distance of 4 meters and 9 meters from the base of the tower and in the same straight line with it are complementary. Find the height of the tower.

Answer: the height of the tower is 6 meters.

Solution: Let the height of the tower be h meters and the two angles of elevation be α and (90° - α).

From the first point, we have:

tan(α) = h / 4

From the second point, we have:

tan(90° - α) = h / 9

Multiply the two equations:

tan(α) × tan(90° - α) = (h / 4) × (h / 9)

tan(α) × cot(α) = h² / 36

1 = h² / 36

Solve for h:

h² = 36

h = √36

h = 6

Therefore, the height of the tower is 6 meters.

Q7. A man standing on the deck of a ship, which is 10 meters above the water, observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°. Calculate the distance of the hill from the ship.

Answer: the distance of the hill from the ship is 5 meters.

Solution: Let the distance of the hill from the ship be d meters and the height of the hill above the water level be h meters.

Step 1: Express h in terms of d using the tangent function and the angle of elevation.

tan(60°) = h / d

h = d × tan(60°)

h = d × √3

Step 2: Express h in terms of d using the tangent function and the angle of depression.

tan(30°) = (h - 10) / d

h - 10 = d × tan(30°)

h = d × tan(30°) + 10

h = d × (1/√3) + 10

Step 3: Equate the two expressions for h and solve for d.

d × √3 = d × (1/√3) + 10

d × (√3 - 1/√3) = 10

d × (2√3/√3) = 10

d × (2/1) = 10

2d = 10

d = 5

Therefore, the distance of the hill from the ship is 5 meters.

Q8. From a point P on the ground, the angle of elevation of the top of a 10-meter tall building is 30°. A flag is hoisted at the top of the building, and the angle of elevation of the top of the flag staff from P is 45°. Find the length of the flag staff.

Answer: the length of the flag staff is approximately 7.32 meters.

Solution: Let the length of the flag staff be L meters and the distance between point P and the base of the building be d meters.

Step 1: Express d in terms of the building's height using the tangent function and the angle of elevation of the top of the building.

tan(30°) = 10 / d

d = 10 / tan(30°)

d = 10 × √3

Step 2: Express the total height (building + flag staff) in terms of d using the tangent function and the angle of elevation of the top of the flag staff.

tan(45°) = (10 + L) / d

d × tan(45°) = 10 + L

10√3 × 1 = 10 + L

10√3 = 10 + L

Step 3: Solve for L.

L = 10√3 - 10

L = 10(√3 - 1)

L ≈ 7.32

Therefore, the length of the flag staff is approximately 7.32 meters.

## What were the previous year's CAT Heights & Distance questions?

Q9. The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the jet plane is flying at a constant height of 1500√3 meters, find the speed of the jet plane.

Answer: the speed of the jet plane is 200 meters per second or 720 kilometers per hour.

Solution: Let the distance traveled by the jet plane in 15 seconds be d meters.

Step 1: Express the distances from point A to the jet plane at the two given times using the tangent function.

Initial distance = 1500√3 / tan(60°) = 1500√3 × (√3/3) = 1500 meters

Final distance = 1500√3 / tan(30°) = 1500√3 × √3 = 4500 meters

Step 2: Calculate the distance traveled by the jet plane in 15 seconds.

d = Final distance - Initial distance

d = 4500 - 1500 = 3000 meters

Step 3: Calculate the speed of the jet plane in meters per second.

Speed = Distance / Time

Speed = 3000 / 15 = 200 meters per second

Step 4: Convert the speed to kilometers per hour.

Speed in km/h = (Speed in m/s) × 3600 / 1000

Speed in km/h = 200 × 3600 / 1000 = 720 km/h

Therefore, the speed of the jet plane is 200 meters per second or 720 kilometers per hour.

Q10. 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 is 60°. When the person moves 40 meters away from the bank, the angle of elevation becomes 30°. Find the height of the tree.

Answer: the height of the tree is approximately 34.64 meters.

Solution: Let the height of the tree be h meters and the initial distance of the person from the base of the tree be d meters.

Step 1: Express h in terms of d using the tangent function and the initial angle of elevation.

tan(60°) = h / d

h = d × tan(60°)

h = d × √3

Step 2: Express h in terms of (d + 40) using the tangent function and the new angle of elevation.

tan(30°) = h / (d + 40)

h = (d + 40) × tan(30°)

h = (d + 40) × (1/√3)

Step 3: Equate the two expressions for h and solve for d.

d × √3 = (d + 40) × (1/√3)

d × √3 × √3 = (d + 40)

3d = d + 40

2d = 40

d = 20

Step 4: Calculate the height of the tree by substituting the value of d in either of the expressions for h.

h = d × √3

h = 20 × √3

h = 20√3

h ≈ 34.64

Therefore, the height of the tree is approximately 34.64 meters.

Q11. From the top of a building, the angles of depression of the top and bottom of a vertical lamp post are 30° and 60°, respectively. If the lamp post is 10 meters tall, find the height of the building.

Answer: the height of the building is approximately 3.33 meters.

Solution: Let the height of the building be h meters and the distance between the building and the lamp post be d meters.

Step 1: Express d in terms of the lamp post's height using the tangent function and the angle of depression of the bottom of the lamp post.

tan(60°) = 10 / d

d = 10 / tan(60°)

d = 10 / √3

Step 2: Express (h + 10) in terms of d using the tangent function and the angle of depression of the top of the lamp post.

tan(30°) = h / d

h = d × tan(30°)

h = (10 / √3) × (1/√3)

h = 10 / 3

Step 3: Solve for the height of the building.

Height of the building = h

Height of the building = 10 / 3

Height of the building ≈ 3.33

Therefore, the height of the building is approximately 3.33 meters.

Q12. A statue, 1.6 meters tall, stands on the top of a pedestal. From a point on the ground, the angles of elevation of the top of the statue and the top of the pedestal are 60° and 45°, respectively. Find the height of the pedestal.

Answer: the height of the pedestal is approximately 3.52 meters.

Solution: Sure! Here's the direct solution to find the height of the pedestal.

Given:

The statue is 1.6 meters tall.

From a point on the ground, the angle of elevation of the top of the statue is 60°, and the angle of elevation of the top of the pedestal is 45°.

Let the height of the pedestal be h meters and the distance between the point on the ground and the base of the pedestal be d meters.

Step 1: Express (h + 1.6) in terms of d using the tangent function and the angle of elevation of the top of the statue.

tan(60°) = (h + 1.6) / d

d × tan(60°) = h + 1.6

d × √3 = h + 1.6

Step 2: Express h in terms of d using the tangent function and the angle of elevation of the top of the pedestal.

tan(45°) = h / d

h = d × tan(45°)

h = d × 1

h = d

Step 3: Equate the two expressions and solve for d.

d × √3 = d + 1.6

d(√3 - 1) = 1.6

d = 1.6 / (√3 - 1)

Step 4: Calculate the height of the pedestal by substituting the value of d in the expression for h.

h = d

h = 1.6 / (√3 - 1)

h ≈ 3.52

Therefore, the height of the pedestal is approximately 3.52 meters.

## What were the Heights & Distance questions on the CAT 2022 exam?

Q13. Two poles of equal heights are standing opposite each other on either side of the road, which is 80 meters wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles.

Answer: the height of the poles is approximately 27.72 meters.

Solution: Excellent! Let's solve this problem step by step to find the height of the poles.
Given:

Two poles of equal heights are standing opposite each other on either side of the road.

The road is 80 meters wide.

From a point between the poles on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively.

Let the height of the poles be h meters and the distance of the point from the pole with a 30° angle of elevation be d meters.

Step 1: Express the width of the road in terms of d using the tangent function and the given angles of elevation.

tan(60°) = h / d

tan(30°) = h / (80 - d)

Step 2: Divide the two equations to eliminate h.

tan(60°) / tan(30°) = (h / d) / (h / (80 - d))

√3 = (80 - d) / d

Step 3: Solve for d.

d × √3 = 80 - d

d × √3 + d = 80

d(√3 + 1) = 80

d = 80 / (√3 + 1)

Step 4: Calculate the height of the poles by substituting the value of d in either of the expressions for h.

tan(30°) = h / (80 - d)

h = (80 - d) × tan(30°)

h = (80 - (80 / (√3 + 1))) × (1/√3)

h = (80 - (80 / (√3 + 1))) / √3

h = (80√3 - 80) / (√3 + 1)

h = 80(√3 - 1) / (√3 + 1)

h ≈ 27.72

Therefore, the height of the poles is approximately 27.72 meters.

Q14. A 1.2-meter tall girl spots a balloon moving with the wind in a straight line at a height of 88.2 meters from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance traveled by the balloon during this time.

Answer: the distance traveled by the balloon during this time is approximately 100.4 meters.

Solution: Let the initial distance of the balloon from the girl be d1 meters and the final distance be d2 meters.

Step 1: Express the height of the balloon from the girl's eyes in terms of d1 and d2 using the tangent function.

Height of the balloon from the girl's eyes = 88.2 - 1.2 = 87 meters

tan(60°) = 87 / d1

d1 = 87 / tan(60°)

d1 = 87 / √3

tan(30°) = 87 / d2

d2 = 87 / tan(30°)

d2 = 87 × √3

Step 2: Calculate the distance traveled by the balloon.

Distance traveled = d2 - d1

Distance traveled = 87 × √3 - 87 / √3

Distance traveled = 87(√3 - 1/√3)

Distance traveled = 87 × (√3)² - 1 / √3

Distance traveled = 87 × 2 / √3

Distance traveled = 174 / √3

Distance traveled ≈ 100.4

Therefore, the distance traveled by the balloon during this time is approximately 100.4 meters.

Q15. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Answer: the car will take 3 seconds to reach the foot of the tower from the point where the angle of depression is 60°.

Solution: Sure! Let's solve this problem step by step to find the time taken by the car to reach the foot of the tower.

Given:

A man standing at the top of a tower observes a car approaching the foot of the tower with a uniform speed.

Initially, the angle of depression of the car is 30°.

After six seconds, the angle of depression of the car is 60°.

Let the height of the tower be h meters, and the distance of the car from the foot of the tower at the initial and final positions be d1 and d2 meters, respectively.

Step 1: Express the distances d1 and d2 in terms of the height of the tower using the tangent function.

tan(30°) = h / d1

d1 = h × cot(30°)

d1 = h × √3

tan(60°) = h / d2

d2 = h × cot(60°)

d2 = h / √3

Step 2: Calculate the distance traveled by the car in 6 seconds.

Distance traveled in 6 seconds = d1 - d2

Distance traveled in 6 seconds = h × √3 - h / √3

Distance traveled in 6 seconds = h(√3 - 1/√3)

Distance traveled in 6 seconds = h × 2/√3

Step 3: Let the time taken by the car to reach the foot of the tower from the final position be t seconds. Set up an equation using the distance-speed-time formula.

Distance traveled in 6 seconds / 6 = d2 / t

(h × 2/√3) / 6 = (h / √3) / t

Step 4: Solve for t.

2h / (6√3) = h / (t√3)

2t = 6

t = 3

Therefore, the car will take 3 seconds to reach the foot of the tower from the point where the angle of depression is 60°.

Q16. The top of a building is 60 meters above the ground. From the top, the angles of depression of the top and bottom of another building are observed to be 30° and 60°, respectively. Find the height of the other building.

Answer: the height of the other building is 40 meters.

Solution: Let the height of the other building be h meters and the distance between the two buildings be d meters.

Step 1: Express the distance d in terms of the height of the first building using the tangent function and the angle of depression of the bottom of the other building.

tan(60°) = 60 / d

d = 60 × cot(60°)

d = 60 / √3

Step 2: Express the height of the other building in terms of the distance d using the tangent function and the angle of depression of the top of the other building.

tan(30°) = (60 - h) / d

h = 60 - d × tan(30°)

h = 60 - (60 / √3) × (1/√3)

h = 60 - 60/3

h = 60 × 2/3

h = 40

Therefore, the height of the other building is 40 meters.

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