Question 1

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

Accepted Answer

Step 1: We have a right-angled triangle where the horizontal distance (base) = 30 m and angle of elevation = 60°. Step 2: Using tan(60°) = height/base, we get √3 = h/30. Step 3: h = 30√3 metres ≈ 51.96 metres. Since 30√3 is the exact form, the answer is 30√3 m.

Question 2

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: In this right-angled triangle, base = 40 m and angle of elevation = 45°. Step 2: Using tan(45°) = height/base, we get 1 = h/40. Step 3: h = 40 metres.

Question 3

An observer on the ground sees the top of a tree at an angle of elevation of 30°. If the observer is 50√3 metres away from the base of the tree, what is the height of the tree?

Accepted Answer

Step 1: Base distance = 50√3 m, angle of elevation = 30°. Step 2: Using tan(30°) = height/base, we get 1/√3 = h/(50√3). Step 3: h = (50√3) × (1/√3) = 50 metres.

Question 4

From the top of a 20-metre tall building, the angle of depression to an object on the ground is 30°. Find the horizontal distance from the building to the object.

Accepted Answer

Step 1: Height of building = 20 m. Angle of depression = 30°. Step 2: Angle of depression equals angle of elevation from the object's perspective. Step 3: Using tan(30°) = height/distance, we get 1/√3 = 20/d. Step 4: d = 20√3 metres.

Question 5

A man standing 100 m away from the base of a tower observes the angle of elevation to the top of the tower as 30°. Find the height of the tower (in metres). [Use √3 ≈ 1.732]

Accepted Answer

Step 1: Set up the right triangle where horizontal distance = 100 m, angle of elevation = 30°, height = h. Step 2: Use tan(30°) = h/100. Step 3: tan(30°) = 1/√3, so h = 100/√3 = 100/1.732 ≈ 57.74 m. Step 4: Rounding to nearest integer gives approximately 58 m.

Question 6

From the top of a cliff 80 m high, the angle of depression to a boat on the water is 45°. How far is the boat from the base of the cliff (in metres)?

Accepted Answer

Step 1: Set up the right triangle where height of cliff = 80 m, angle of depression = 45°. Step 2: Angle of depression equals angle of elevation from the boat's perspective. Step 3: Use tan(45°) = height/horizontal distance = 80/d. Step 4: tan(45°) = 1, so d = 80 m.

Question 7

Two buildings stand on level ground. From the top of the first building (height 20 m), the angle of elevation to the top of the second building is 30°. The horizontal distance between the buildings is 40 m. Find the height of the second building (in metres). [Use √3 ≈ 1.732]

Accepted Answer

Step 1: Let the second building's height be H. The vertical difference between tops = (H - 20) m. Step 2: From the top of the first building, tan(30°) = (H - 20)/40. Step 3: tan(30°) = 1/√3, so (H - 20)/40 = 1/√3. Step 4: H - 20 = 40/√3 = 40/1.732 ≈ 23.09. Step 5: H = 20 + 23.09 ≈ 43.1 m.

Question 8

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

Accepted Answer

Step 1: Let the tree height = h, initial distance = d. From first position: tan(45°) = h/d, so h = d (since tan(45°) = 1). Step 2: From second position (10 m closer): tan(60°) = h/(d - 10), so h = √3(d - 10). Step 3: Equate: d = √3(d - 10). Step 4: d = √3·d - 10√3, so d(√3 - 1) = 10√3. Step 5: d = 10√3/(√3 - 1) = 10√3(√3 + 1)/[(√3 - 1)(√3 + 1)] = 10√3(√3 + 1)/2 = 10(3 + √3)/2 = 5(3 + √3) ≈ 5(3 + 1.732) = 23.66 m. Step 6: h = d ≈ 23.66 m ≈ 23.7 m.

Question 9

From a point on the ground, the angle of elevation to the top of a building is 30°. From a point 50 m further away (on the same line), the angle of elevation becomes 15°. Find the height of the building (in metres). [Use √3 ≈ 1.732, tan(15°) ≈ 0.268]

Accepted Answer

Step 1: Let building height = h, initial distance = d. From first position: tan(30°) = h/d, so h = d/√3. Step 2: From second position: tan(15°) = h/(d + 50), so h = 0.268(d + 50). Step 3: Equate: d/√3 = 0.268(d + 50). Step 4: d/1.732 = 0.268d + 13.4. Step 5: 0.578d = 0.268d + 13.4. Step 6: 0.31d = 13.4, so d ≈ 43.23 m. Step 7: h = 43.23/1.732 ≈ 24.96 ≈ 25 m.

Question 10

From the top of a cliff 120 m high, the angles of depression to two boats on the sea are 30° and 45° respectively. If both boats are on the same side of the cliff, find the distance between the boats (in metres).

Accepted Answer

Step 1: Let cliff height = 120 m. For boat 1 (angle 30°): tan(30°) = 120/x₁, so x₁ = 120/tan(30°) = 120√3. Step 2: For boat 2 (angle 45°): tan(45°) = 120/x₂, so x₂ = 120/tan(45°) = 120. Step 3: Distance between boats = x₁ − x₂ = 120√3 − 120 = 120(√3 − 1) ≈ 120(1.732 − 1) = 120(0.732) ≈ 87.8 m.

Question 11

A ladder leans against a vertical wall. The angle between the ladder and the ground is 60°. If the ladder is 10 m long, how far is the base of the ladder from the wall? (in metres)

Accepted Answer

Step 1: The ladder forms the hypotenuse of a right triangle. Length of ladder = 10 m, angle with ground = 60°. Step 2: Distance from wall (adjacent to 60° angle) = ladder × cos(60°) = 10 × (1/2) = 5 m.

Question 12

From a point on the ground 100 m away from the base of a vertical pole, the angle of elevation to the top is 45°. A person climbs the pole and the angle of elevation from the same point becomes 60°. How much did the person climb (in metres)?

Accepted Answer

Step 1: Initial angle of elevation = 45°, distance = 100 m. Height of pole: tan(45°) = h₁/100, so h₁ = 100 m. Step 2: After climbing, angle = 60°. New height: tan(60°) = h₂/100, so h₂ = 100√3 m. Step 3: Distance climbed = h₂ − h₁ = 100√3 − 100 = 100(√3 − 1) ≈ 73.2 m.

SSC CPO Heights & Distances

Heights & Distances— Rules & Concept

Key Points to Remember

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Heights & Distances — Practice Questions

60-Second Revision — Heights & Distances