Question 1

A ladder of length 10 m is leaning against a wall. If the ladder makes an angle of 30° with the ground, what is the height of the point on the wall where the ladder touches?

Accepted Answer

Step 1: We use sin(angle) = opposite/hypotenuse. Step 2: sin(30°) = height/10. Step 3: height = 10 × sin(30°) = 10 × (1/2) = 5 m. Therefore, the height is 5 m.

Question 2

Two buildings are 50 metres apart. From the top of the first building, the angle of depression to the top of the second building is 30°. If the first building is 80 metres tall, what is the height of the second building? (Use √3 = 1.732)

Accepted Answer

Step 1: The angle of depression of 30° means the line of sight goes down at 30° from horizontal. Step 2: The vertical drop = horizontal distance × tan(30°). Step 3: Vertical drop = 50 × (1/√3) = 50/1.732 ≈ 28.87 metres. Step 4: Height of second building = 80 − 28.87 ≈ 51 metres.

Question 3

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? (Use √3 = 1.732)

Accepted Answer

Step 1: Angle of depression = angle of elevation from the boat's perspective. Step 2: tan(30°) = height/horizontal distance. Step 3: 1/√3 = 80/distance. Step 4: distance = 80√3 = 80 × 1.732 = 138.56 ≈ 139 metres.

Question 4

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. (Use √3 = 1.732)

Accepted Answer

Step 1: In a right-angled triangle, tan(angle) = opposite/adjacent. Step 2: Here, tan(60°) = height/30. Step 3: √3 = height/30, so height = 30√3 = 30 × 1.732 = 51.96 metres ≈ 52 metres.

Question 5

From the top of a 25-metre tall tower, an observer sees a car on the ground at an angle of depression of 45°. How far is the car from the base of the tower?

Accepted Answer

Step 1: Use the angle of depression property: tan(45°) = height / horizontal distance. Step 2: tan(45°) = 1, so 1 = 25 / distance. Step 3: distance = 25 metres.

Question 6

A person standing on a bridge observes a boat in the river below at an angle of depression of 30°. If the bridge is 20 metres high, how far is the boat from the point directly below the observer?

Accepted Answer

Step 1: The angle of depression equals the angle of elevation from the boat's perspective. Step 2: tan(30°) = height / horizontal distance = 20 / distance. Step 3: tan(30°) = 1/√3, so 1/√3 = 20 / distance. Step 4: distance = 20√3 metres.

Question 7

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: Use the tangent ratio in the right triangle formed. tan(60°) = height / distance. Step 2: tan(60°) = √3, so √3 = height / 30. Step 3: height = 30√3 metres.

Question 8

An observer on the ground sees the top of a cliff at an angle of elevation of 30°. If the observer moves 50 metres closer to the cliff, the angle of elevation becomes 60°. Find the height of the cliff.

Accepted Answer

Step 1: Let height = h, initial distance = d. From first position: tan(30°) = h/d, so h = d/√3. Step 2: From second position: tan(60°) = h/(d-50), so h = (d-50)√3. Step 3: Equate: d/√3 = (d-50)√3. Step 4: d = 3(d-50), so d = 3d - 150, giving 2d = 150, d = 75. Step 5: h = 75/√3 = 25√3 metres.

Question 9

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, so 1 = height / 40. Step 3: height = 40 metres.

Question 10

From a point on the ground 40 metres away from a building, the angle of elevation to the top is 45°. From the same point, the angle of elevation to a window is 30°. Find the vertical distance between the top and the window (in metres). [Use √3 = 1.732]

Accepted Answer

Step 1: Distance from building = 40 m. Angle to top = 45°, angle to window = 30°. Step 2: Height to top = 40 × tan(45°) = 40 × 1 = 40 m. Step 3: Height to window = 40 × tan(30°) = 40 × (1/√3) = 40/1.732 ≈ 23.09 m. Step 4: Vertical distance = 40 - 23.09 = 16.91 metres.

Question 11

Two buildings of heights 20 metres and 35 metres stand on level ground 15 metres apart. Find the angle of depression from the top of the taller building to the top of the shorter building. [Use tan⁻¹(1) = 45°, tan⁻¹(1/3) ≈ 18.43°]

Accepted Answer

Step 1: Difference in heights = 35 - 20 = 15 metres. Horizontal distance between buildings = 15 metres. Step 2: The angle of depression θ satisfies: tan(θ) = vertical drop / horizontal distance = 15/15 = 1. Step 3: θ = tan⁻¹(1) = 45°.

Question 12

A ladder of length 13 metres leans against a wall. The angle between the ladder and the ground is 67°. How high does the ladder reach on the wall (in metres)? [Use sin(67°) = 0.92]

Accepted Answer

Step 1: Length of ladder = 13 m (hypotenuse). Angle with ground = 67°. Step 2: Height on wall = ladder length × sin(angle with ground) = 13 × sin(67°) = 13 × 0.92 = 11.96 metres.

Question 13

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 (in metres)? [Use √3 = 1.732]

Accepted Answer

Step 1: Height of cliff = 80 m. Angle of depression = 30°. Step 2: Angle of depression equals angle of elevation from boat's perspective. Step 3: Using tan(30°) = height/distance, we get 1/√3 = 80/distance. Step 4: distance = 80√3 = 80 × 1.732 = 138.56 metres.

Question 14

A man standing 100 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 (in metres). [Use √3 = 1.732]

Accepted Answer

Step 1: Let height of tower = h metres. Distance from base = 100 m. Angle of elevation = 60°. Step 2: Using tan(60°) = h/100, we get h = 100 × tan(60°) = 100 × √3 = 100 × 1.732 = 173.2 metres.

Question 15

A man standing 40 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 (in metres). [Use √3 = 1.732]

Accepted Answer

Step 1: Let height of tower = h metres. Distance from base = 40 metres. Angle of elevation = 60°. Step 2: Using tan(60°) = h/40, we get h = 40 × tan(60°) = 40 × √3. Step 3: h = 40 × 1.732 = 69.28 metres ≈ 69.3 metres.

Question 16

Two buildings of heights 25 metres and 40 metres stand on level ground. The angle of depression from the top of the taller building to the top of the shorter building is 30°. A person at the top of the shorter building observes the angle of elevation to the top of the taller building as θ. Find tan(θ).

Accepted Answer

Step 1: The vertical difference between the tops of the buildings is 40 - 25 = 15 metres. Step 2: From the top of the taller building, angle of depression to the top of the shorter building is 30°. This means tan(30°) = 15 / (horizontal distance d), so d = 15 / tan(30°) = 15√3 metres. Step 3: From the top of the shorter building, the angle of elevation θ to the top of the taller building has tan(θ) = 15 / (15√3) = 1/√3 = √3/3.

Question 17

From the top of a cliff 80 metres high, the angle of depression to a boat on the sea is 30°. The boat moves directly away from the cliff. After some time, the angle of depression becomes 15°. How far (in metres) has the boat travelled? (Use √3 ≈ 1.732)

Accepted Answer

Step 1: Let the initial horizontal distance of the boat from the cliff be d₁. From the top of the cliff, tan(30°) = 80/d₁, so d₁ = 80/tan(30°) = 80/(1/√3) = 80√3 metres. Step 2: After the boat moves, let the new distance be d₂. tan(15°) = 80/d₂, so d₂ = 80/tan(15°). Step 3: tan(15°) = tan(45° - 30°) = (tan(45°) - tan(30°))/(1 + tan(45°)tan(30°)) = (1 - 1/√3)/(1 + 1/√3) = (√3 - 1)/(√3 + 1). Rationalizing: [(√3 - 1)/(√3 + 1)] × [(√3 - 1)/(√3 - 1)] = (√3 - 1)²/(3 - 1) = (3 - 2√3 + 1)/2 = (4 - 2√3)/2 = 2 - √3. Step 4: d₂ = 80/(2 - √3). Rationalizing: 80(2 + √3)/[(2 - √3)(2 + √3)] = 80(2 + √3)/(4 - 3) = 80(2 + √3) = 160 + 80√3 metres. Step 5: Distance travelled = d₂ - d₁ = (160 + 80√3) - 80√3 = 160 metres.

Question 18

A man standing at point A observes the angle of elevation to the top of a tower at point C as 30°. He walks 40√3 metres towards the tower and observes the angle of elevation as 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 the tower above eye level be h metres. From the first position, tan(30°) = h / (horizontal distance d₁), so d₁ = h√3. Step 2: From the second position (40√3 metres closer), tan(60°) = h / (d₁ - 40√3), so √3 = h / (d₁ - 40√3). Step 3: Substituting d₁ = h√3: √3(h√3 - 40√3) = h, which gives 3h - 120 = h, so h = 60 metres. Step 4: Total tower height = 60 + 1.5 = 61.5 metres.

SSC CHSL Heights & Distances

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Heights & Distances — Practice Questions

60-Second Revision — Heights & Distances