Question 1

From the top of a cliff 80 metres high, the angle of depression to a boat on the water is 30°. How far is the boat from the base of the cliff?

Accepted Answer

Step 1: Use the tangent ratio with angle of depression. tan(30°) = height / horizontal distance. Step 2: tan(30°) = 1/√3. Step 3: 1/√3 = 80 / distance. Step 4: distance = 80√3 metres ≈ 138.56 metres. Therefore, the answer is 80√3 metres.

Question 2

A ladder leans against a wall such that it makes an angle of 45° with the ground. If the ladder is 10√2 metres long, what is the height at which the ladder touches the wall?

Accepted Answer

Step 1: Use the sine ratio. sin(45°) = height / hypotenuse (ladder length). Step 2: sin(45°) = 1/√2. Step 3: 1/√2 = height / 10√2. Step 4: height = (10√2) × (1/√2) = 10 metres. Therefore, the answer is 10 metres.

Question 3

A person standing on the ground observes the angle of elevation to the top of a tree to be 60°. If the person moves 10 metres closer to the tree, the angle of elevation becomes 90°. What is the height of the tree?

Accepted Answer

Step 1: When the angle of elevation is 90°, the person is directly at the base of the tree (distance = 0). Step 2: Let the initial distance be d metres. From the initial position: tan(60°) = height / d, so height = d√3. Step 3: After moving 10 metres closer, the person is at distance (d − 10) metres. Since the angle becomes 90°, we have d − 10 = 0, so d = 10 metres. Step 4: Height = 10√3 metres ≈ 17.32 metres. Therefore, the answer is 10√3 metres.

Question 4

From a point on the ground 40 metres away from the base of a building, the angle of elevation to the top is 45°. What is the height of the building?

Accepted Answer

Step 1: Use the tangent ratio. tan(45°) = height / distance. Step 2: tan(45°) = 1. Step 3: 1 = height / 40. Step 4: height = 40 metres. Therefore, the answer is 40 metres.

Question 5

A man standing 30 metres away from the base of a tower observes the angle of elevation to the top of the tower to be 60°. Find the height of the tower.

Accepted Answer

Step 1: Use the tangent ratio in the right triangle formed. tan(60°) = height / distance. Step 2: tan(60°) = √3. Step 3: √3 = height / 30. Step 4: height = 30√3 metres ≈ 51.96 metres. Therefore, the answer is 30√3 metres.

Question 6

A person standing on the ground observes the angle of elevation to the top of a tree as 45°. After walking 10 metres towards the tree, the angle of elevation becomes 60°. Find the height of the tree. [Use √3 ≈ 1.732]

Accepted Answer

Step 1: Let height = h, initial distance = d. From first position: tan(45°) = h/d, so h = d. Step 2: From second position (10 m closer): tan(60°) = h/(d-10), so √3 = h/(d-10). Step 3: Substituting h = d: √3 = d/(d-10). Step 4: √3(d-10) = d, so √3·d - 10√3 = d. Step 5: d(√3 - 1) = 10√3. Step 6: d = 10√3/(√3-1) = 10√3(√3+1)/[(√3-1)(√3+1)] = 10(3+√3)/2 = 5(3+√3) ≈ 23.66 m. Step 7: h = d ≈ 23.66 metres.

Question 7

From the top of a lighthouse 75 m high, the angle of depression to a ship is 30°. The ship moves directly towards the lighthouse. When the ship reaches a point where the angle of depression is 45°, how much closer is the ship to the lighthouse compared to its initial position? [Use √3 ≈ 1.732]

Accepted Answer

Step 1: Let the initial distance of the ship from the base of the lighthouse = d₁. From the top of the lighthouse, angle of depression = 30°, so tan(30°) = 75/d₁, giving d₁ = 75/tan(30°) = 75√3 m. Step 2: When the angle of depression becomes 45°, let the new distance = d₂. Then tan(45°) = 75/d₂, giving d₂ = 75/tan(45°) = 75/1 = 75 m. Step 3: Distance moved closer = d₁ − d₂ = 75√3 − 75 = 75(√3 − 1) m ≈ 75(1.732 − 1) = 75 × 0.732 = 54.9 m ≈ 75(√3−1) m.

Question 8

A person standing on the ground observes the angle of elevation to the top of a building as 45°. He walks 20 m towards the building and observes the angle of elevation as 60°. Find the height of the building. [Use √3 ≈ 1.732]

Accepted Answer

Step 1: Let the initial distance from the building = x m, and height = h m. From the first position: tan(45°) = h/x, so h = x (since tan(45°) = 1). Step 2: From the second position (20 m closer): tan(60°) = h/(x−20), so h = (x−20)√3 (since tan(60°) = √3). Step 3: Equate the two expressions for h: x = (x−20)√3. Step 4: x = x√3 − 20√3, so x(√3−1) = 20√3, giving x = 20√3/(√3−1). Step 5: Rationalize: x = 20√3(√3+1)/[(√3−1)(√3+1)] = 20√3(√3+1)/(3−1) = 20√3(√3+1)/2 = 10√3(√3+1) = 10(3+√3) = 30 + 10√3 m. Step 6: Height h = x = 30 + 10√3 ≈ 30 + 17.32 = 47.32 m. For clean answer: h = 10(3+√3) m or approximately 47.3 m. Alternatively, if we want integer: assume h = 50 m (close approximation).

Question 9

A ladder leans against a vertical wall. The angle between the ladder and the ground is 60°. If the ladder is 8 m long, how far is the base of the ladder from the wall? [Use √3 ≈ 1.732]

Accepted Answer

Step 1: The ladder forms the hypotenuse of a right triangle. The angle between the ladder and the ground is 60°. Step 2: The distance from the base of the ladder to the wall is the side adjacent to the 60° angle. Step 3: Using cos(60°) = adjacent/hypotenuse: cos(60°) = distance/8. Step 4: cos(60°) = 1/2, so distance = 8 × (1/2) = 4 m.

IBPS Clerk Heights & Distances

Heights & Distances— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Heights & Distances — Practice Questions

60-Second Revision — Heights & Distances