Question 1

From the top of a 30-metre tall building, the angle of depression to a point on the ground is 45°. How far is the point from the base of the building?

Accepted Answer

Step 1: Angle of depression equals angle of elevation in the right triangle formed. Step 2: tan(45°) = height/distance. Step 3: tan(45°) = 1. Step 4: 1 = 30/distance. Step 5: distance = 30 metres.

Question 2

A person standing 15√3 metres away from a tower observes the angle of elevation to the top as 30°. What is the height of the tower?

Accepted Answer

Step 1: Use tan(30°) = height/distance. Step 2: tan(30°) = 1/√3. Step 3: height = 15√3 × (1/√3) = 15 metres.

Question 3

An observer on the ground sees the top of a tree at an angle of elevation of 60°. If the observer is 20 metres away from the base of the tree, what is the height of the tree?

Accepted Answer

Step 1: Use tan(60°) = height/distance. Step 2: tan(60°) = √3. Step 3: height = 20 × tan(60°) = 20 × √3 = 20√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 tan(45°) = height/distance. Step 2: tan(45°) = 1. Step 3: height = 40 × tan(45°) = 40 × 1 = 40 metres.

Question 5

A man observes the angle of depression to a point on the ground from the top of a 30-metre building as 45°. How far is the point from the base of the building?

Accepted Answer

Step 1: Angle of depression = angle of elevation from the ground point. Step 2: tan(45°) = height / horizontal distance. Step 3: tan(45°) = 1. Step 4: 1 = 30/d. Step 5: d = 30 metres.

Question 6

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

Accepted Answer

Step 1: tan(30°) = height / horizontal distance. Step 2: tan(30°) = 1/√3. Step 3: 1/√3 = 80/d. Step 4: d = 80√3 metres ≈ 138.56 metres.

Question 7

A man standing at point A observes the angle of elevation to the top of a tower at 30°. He walks 40√3 metres towards the base of the tower and observes the angle of elevation to be 60°. If the man's eye level is 1.5 metres above ground, find the height of the tower (in metres).

Accepted Answer

Step 1: Let the height of tower above eye level = h. From first position: tan(30°) = h / d, where d is initial horizontal distance. So h = d/√3. Step 2: From second position after walking 40√3 m closer: tan(60°) = h / (d - 40√3). So h = √3(d - 40√3). Step 3: Equating: d/√3 = √3(d - 40√3) → d/√3 = √3·d - 120 → d = 3d - 120√3 → 2d = 120√3 → d = 60√3. Step 4: h = 60√3/√3 = 60 metres. Step 5: Total height = 60 + 1.5 = 61.5 metres.

Question 8

From the top of a cliff 120 metres high, the angles of depression to two boats on the water are 30° and 60° respectively. Both boats are on the same side of the cliff. Find the distance between the two boats (in metres).

Accepted Answer

Step 1: Let the cliff height = 120 m. For boat 1 (angle of depression 30°): tan(30°) = 120/d₁ → d₁ = 120/tan(30°) = 120/(1/√3) = 120√3 metres. Step 2: For boat 2 (angle of depression 60°): tan(60°) = 120/d₂ → d₂ = 120/tan(60°) = 120/√3 = 40√3 metres. Step 3: Distance between boats = d₁ - d₂ = 120√3 - 40√3 = 80√3 metres ≈ 138.56 metres.

Question 9

A person standing on the ground observes the angle of elevation to the top of a building as 45°. After walking 50 metres towards the building, the angle of elevation becomes 60°. If the person's eye level is at 1.6 metres, find the height of the building (in metres). [Use √3 ≈ 1.732]

Accepted Answer

Step 1: Let height above eye level = h, initial distance = d. From first position: tan(45°) = h/d → h = d (since tan(45°) = 1). Step 2: From second position: tan(60°) = h/(d-50) → h = √3(d-50). Step 3: Equating: d = √3(d-50) → d = √3·d - 50√3 → 50√3 = √3·d - d → 50√3 = d(√3 - 1). Step 4: d = 50√3/(√3-1) = 50√3(√3+1)/((√3-1)(√3+1)) = 50√3(√3+1)/(3-1) = 50√3(√3+1)/2 = 50(3+√3)/2 = 25(3+√3) = 75 + 25√3 ≈ 75 + 43.3 = 118.3 metres. Step 5: h = d = 118.3 metres. Step 6: Total height = 118.3 + 1.6 = 119.9 ≈ 120 metres.

SBI PO Heights & Distances

Heights & Distances— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Heights & Distances — Practice Questions

60-Second Revision — Heights & Distances