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 to be 60°. Find the height of the tower. (Use √3 ≈ 1.732)

Accepted Answer

Step 1: Let the height of the tower be h metres. The man is standing 30 m away from the base (horizontal distance = 30 m). Step 2: Using the tangent ratio: tan(60°) = h/30. Step 3: Since tan(60°) = √3, we have √3 = h/30. Step 4: Therefore, h = 30√3 = 30 × 1.732 = 51.96 ≈ 52 metres.

Question 2

A man standing 40 metres away from the base of a tower observes that the angle of elevation to the top of the tower is 30°. Find the height of the tower. (Use √3 ≈ 1.732)

Accepted Answer

Step 1: In the right triangle formed, we have the horizontal distance (adjacent) = 40 m and angle of elevation = 30°. Step 2: Using tan(30°) = height/distance, we get height = distance × tan(30°). Step 3: tan(30°) = 1/√3. Step 4: Height = 40 × (1/√3) = 40/√3. Step 5: Rationalizing: (40/√3) × (√3/√3) = 40√3/3 = (40 × 1.732)/3 = 69.28/3 ≈ 23.09 m. Step 6: Rounding to nearest practical value: approximately 23.1 m or 40/√3 m exactly.

Question 3

From the top of a lighthouse 60 m high, the angles of depression of two ships on the same side are 45° and 30°. What is the distance between the two ships?

Accepted Answer

Step 1: Let the lighthouse be AB = 60 m. For the nearer ship C, angle of depression = 45°. tan 45° = 60/BC → BC = 60 m. Step 2: For the farther ship D, angle of depression = 30°. tan 30° = 60/BD → BD = 60√3 m. Step 3: Distance between ships = BD − BC = 60√3 − 60 = 60(√3 − 1) m. Using √3 ≈ 1.732: 60 × 0.732 = 43.92 m. In exact form: 60(√3 − 1) m. Therefore, the answer is 60(√3 − 1) m.

Question 4

From a point on the ground 40 m away from the base of a vertical tower, the angle of elevation to the top is 60°. A man climbs to a point on the tower such that the angle of elevation from the same ground point becomes 45°. How many metres did the man climb?

Accepted Answer

Step 1: Let the height of the tower be H. From the ground point 40 m away, tan(60°) = H/40, so H = 40√3 m. Step 2: Let the man climb to height h. From the same ground point, tan(45°) = h/40, so h = 40 m. Step 3: Distance climbed = H − h = 40√3 − 40 = 40(√3 − 1) m ≈ 40(1.732 − 1) = 40(0.732) = 29.28 m. Rounding to nearest clean value: 40(√3 − 1) ≈ 29.3 m, or exactly 40(√3 − 1) m.

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 as 60°. Find the height of the tower.

Accepted Answer

Step 1: We have a right-angled triangle where the horizontal distance (adjacent) = 30 m and angle of elevation = 60°. Step 2: Using tan(60°) = height/distance, we get height = 30 × tan(60°) = 30 × √3 = 30√3 metres ≈ 51.96 metres. Step 3: Rounding to standard form, height = 30√3 m.

Question 6

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: The ladder forms the hypotenuse of a right-angled triangle. The angle with the ground is 45°. Step 2: Using sin(45°) = height/hypotenuse, we get height = 10√2 × sin(45°). Step 3: sin(45°) = 1/√2, so height = 10√2 × (1/√2) = 10 metres.

Question 7

A ladder leans against a wall such that its foot is 6 metres away from the wall. The angle of elevation from the foot of the ladder to the top of the wall is 45°. What is the height of the wall?

Accepted Answer

Step 1: In the right triangle, the horizontal distance (base) = 6 metres, angle of elevation = 45°. Step 2: tan(45°) = height / base. Step 3: 1 = height / 6. Step 4: height = 6 metres.

Question 8

A vertical pole of height 15 metres casts a shadow of 15 metres on the ground. What is the angle of elevation of the sun?

Accepted Answer

Step 1: Use tan(θ) = opposite/adjacent = height/shadow. Step 2: tan(θ) = 15/15 = 1. Step 3: When tan(θ) = 1, θ = 45°.

Question 9

An observer standing on the ground sees the top of a tree at an angle of elevation of 30°. If the observer moves 10 metres closer to the tree, the angle of elevation becomes 60°. Find the height of the tree.

Accepted Answer

Step 1: Let height = h, initial distance = d. Step 2: From first position: tan(30°) = h/d, so h = d/√3. Step 3: From second position: tan(60°) = h/(d-10), so h = (d-10)√3. Step 4: Equate: d/√3 = (d-10)√3. Step 5: d = 3(d-10). Step 6: d = 3d - 30, so 2d = 30, d = 15. Step 7: h = 15/√3 = 5√3 metres.

Question 10

From a point on the ground, the angle of elevation to the top of a 20-metre tall building is 45°. How far is the point from the base of the building?

Accepted Answer

Step 1: Use tan(θ) = opposite/adjacent. Step 2: tan(45°) = 20/distance. Step 3: 1 = 20/distance. Step 4: distance = 20 metres.

Question 11

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? (Use √3 = 1.732)

Accepted Answer

Step 1: In the right triangle, the height of the cliff is the opposite side = 80 m. 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 12

A man standing 30 metres away from the base of a tower observes that the angle of elevation to the top of the tower is 60°. Find the height of the tower. (Use √3 = 1.732)

Accepted Answer

Step 1: We use the tangent ratio in the right triangle formed. tan(60°) = height / distance. Step 2: tan(60°) = √3, so √3 = h / 30. Step 3: h = 30√3 = 30 × 1.732 = 51.96 metres ≈ 52 metres.

Question 13

From the top of a 40-metre-high cliff, a person observes a boat in the sea at an angle of depression of 30°. How far is the boat from the base of the cliff?

Accepted Answer

Step 1: The angle of depression is 30°, so the angle of elevation from the boat to the top of the cliff is also 30°. Step 2: In the right triangle, height = 40 m (opposite), distance = ? (adjacent). Step 3: tan(30°) = opposite/adjacent = 40/distance. Step 4: Since tan(30°) = 1/√3, we have 1/√3 = 40/distance. Step 5: distance = 40√3 metres.

Question 14

A person standing on the roof of a 20-metre-high building observes the angle of depression to a point on the ground to be 45°. What is the horizontal distance from the building to that point?

Accepted Answer

Step 1: The angle of depression is 45°, which equals the angle of elevation from the ground point to the roof. Step 2: In the right triangle formed, the height is 20 m (opposite side) and the horizontal distance is the adjacent side. Step 3: tan(45°) = opposite/adjacent = 20/distance. Step 4: Since tan(45°) = 1, we have 1 = 20/distance, so distance = 20 metres.

Question 15

An observer on the ground sees the top of a tree at an angle of elevation of 30°. If the observer moves 20 metres closer to the tree, the angle of elevation becomes 60°. Find the height of the tree.

Accepted Answer

Step 1: Let height = h, initial distance = d. Step 2: tan(30°) = h/d → h = d/√3. Step 3: tan(60°) = h/(d-20) → h = √3(d-20). Step 4: Equate: d/√3 = √3(d-20). Step 5: d = 3(d-20) → d = 3d - 60 → 2d = 60 → d = 30. Step 6: h = 30/√3 = 10√3 metres.

Question 16

From a point on the ground 50 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(θ) = opposite/adjacent. Step 2: tan(45°) = height/50. Step 3: Since tan(45°) = 1, we have height = 50 × 1 = 50 metres.

Question 17

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: The angle of depression from the top equals the angle of elevation from the boat's position (alternate angles). Step 2: We have tan(30°) = height/horizontal distance = 80/distance. Step 3: tan(30°) = 1/√3, so 1/√3 = 80/distance. Step 4: distance = 80√3 metres ≈ 138.56 metres.

Question 18

Two buildings are 50 metres apart. From the top of the first building (height 30 m), the angle of depression to the top of the second building is 15°. What is the height of the second building? (Use tan(15°) ≈ 0.27)

Accepted Answer

Step 1: The angle of depression from the top of the first building to the top of the second building is 15°. Step 2: This creates a right-angled triangle where the horizontal distance is 50 m and the vertical drop is (30 - h₂), where h₂ is the height of the second building. Step 3: tan(15°) = (30 - h₂)/50. Step 4: 0.27 = (30 - h₂)/50, so 30 - h₂ = 13.5, thus h₂ = 16.5 metres.

Question 19

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: We use the trigonometric ratio tan(θ) = opposite/adjacent. Step 2: Here, tan(60°) = height/30. Step 3: Since tan(60°) = √3, we have height = 30√3 metres. Step 4: 30√3 ≈ 30 × 1.732 = 51.96 ≈ 52 metres (or exact answer 30√3 m).

Question 20

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

Accepted Answer

Step 1: Use tan(θ) = opposite/adjacent for angle of depression. Step 2: tan(30°) = 80/distance. Step 3: 1/√3 = 80/distance. Step 4: distance = 80√3 metres.

Question 21

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 trigonometric ratio tan(θ) = opposite/adjacent. Step 2: tan(60°) = height/30. Step 3: √3 = height/30. Step 4: height = 30√3 metres ≈ 51.96 metres.

Question 22

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

Accepted Answer

Step 1: Ladder length = 10 m (hypotenuse). Angle with ground = 60°. Step 2: The distance from wall to base of ladder is the adjacent side to the 60° angle. Step 3: cos(60°) = adjacent/hypotenuse = distance/10. Step 4: cos(60°) = 1/2, so distance = 10 × 1/2 = 5 metres.

Question 23

From the top of a cliff 80 metres high, the angle of depression to a boat on the water is 45°. What is the horizontal distance of the boat from the base of the cliff?

Accepted Answer

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

Question 24

From a point on the ground, the angle of elevation to the top of a tree is 45°. If the observer moves 10 m closer to the tree, the angle of elevation becomes 60°. Find the height of the tree (in metres). [Use √3 ≈ 1.732]

Accepted Answer

Step 1: Let initial distance = d, height = h. From first position: tan(45°) = h/d → h = d (since tan(45°) = 1). Step 2: From second position: tan(60°) = h/(d-10) → h = √3(d-10). Step 3: Equating: d = √3(d-10) → d = √3·d - 10√3 → d(√3 - 1) = 10√3 → d = 10√3/(√3-1). Step 4: Rationalizing: d = 10√3(√3+1)/[(√3-1)(√3+1)] = 10√3(√3+1)/2 = 10(3+√3)/2 = 5(3+√3) = 15 + 5√3 ≈ 23.66 m. Step 5: h = d ≈ 23.66 m ≈ 24 m (or exactly 15 + 5√3 m).

IBPS RRB Clerk Heights & Distances

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