Question 1

From the top of a 45-metre-high building, the angle of depression to a point on the ground is 30°. How far is the point from the base of the building? (Use √3 = 1.732)

Accepted Answer

Step 1: In the right triangle formed, tan(30°) = height / horizontal distance. Step 2: tan(30°) = 1/√3, so 1/√3 = 45 / d. Step 3: d = 45√3 = 45 × 1.732 = 77.94 ≈ 78 metres.

Question 2

A man standing at point A observes the angle of elevation to the top of a flagpole as 30°. He moves 20 metres towards the pole to point B, where the angle of elevation becomes 45°. What is the height of the flagpole?

Accepted Answer

Step 1: From position A: tan(30°) = h/d, so h = d/√3. Step 2: From position B: tan(45°) = h/(d-20), so h = d-20. Step 3: Equating: d/√3 = d-20. Step 4: d = √3(d-20) gives d ≈ 47.3 metres. Step 5: h = d-20 ≈ 27 metres.

Question 3

From a point on the ground 50 metres away from the base of a tower, the angle of elevation to the top is 45°. What is the height of the tower?

Accepted Answer

Step 1: Use tan(45°) = height / distance. Step 2: tan(45°) = 1, so 1 = h / 50. Step 3: h = 50 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: Use the tangent ratio: 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 5

From a point on the ground 40 metres away from the base of a building, the angle of elevation to the top is 45°. Another person standing on the roof of the building observes a car on the ground at an angle of depression of 60°. What is the horizontal distance of the car from the base of the building?

Accepted Answer

Step 1: First, find the height of the building. Using tan(45°) = height/40, we get height = 40 × tan(45°) = 40 × 1 = 40 metres. Step 2: Now, from the roof, the angle of depression to the car is 60°. This means tan(60°) = height/horizontal distance of car. Step 3: 40/distance = tan(60°) = √3. Step 4: distance = 40/√3 = 40√3/3 ≈ 23.09 metres. Rationalizing: 40/√3 = 40√3/3 metres.

Question 6

A ladder of length 25 metres leans against a wall. The angle between the ladder and the ground is 60°. How far is the base of the ladder from the wall?

Accepted Answer

Step 1: The ladder forms the hypotenuse of a right-angled 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 adjacent side to the 60° angle. Step 3: Using cos(60°) = adjacent/hypotenuse, we get distance = 25 × cos(60°) = 25 × (1/2) = 12.5 metres.

Question 7

Two buildings of heights 30 metres and 50 metres stand on level ground. The angle of elevation from the top of the shorter building to the top of the taller building is 30°. What is the horizontal distance between the two buildings?

Accepted Answer

Step 1: The difference in heights = 50 - 30 = 20 metres. This is the vertical distance we need to consider. Step 2: From the top of the shorter building, the angle of elevation to the top of the taller building is 30°. Step 3: Using tan(30°) = vertical difference/horizontal distance, we get tan(30°) = 20/distance. Step 4: 1/√3 = 20/distance → distance = 20√3 metres.

Question 8

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

Accepted Answer

Step 1: The angle of depression from the top of the cliff to the boat is 30°. This means the angle of elevation from the boat to the top of the cliff is also 30°. Step 2: We have a right-angled triangle with height = 80 m and angle = 30°. Step 3: Using tan(30°) = height/horizontal distance, we get tan(30°) = 80/distance. Step 4: 1/√3 = 80/distance → distance = 80√3 metres.

Question 9

Two vertical poles of heights 15 metres and 10 metres stand 12 metres apart on level ground. A rope connects the top of the first pole to the top of the second pole. Find the length of the rope (in metres).

Accepted Answer

Step 1: The two poles have heights 15m and 10m, separated horizontally by 12m. Step 2: The difference in heights = 15 − 10 = 5m. Step 3: Consider the rope as the hypotenuse of a right triangle where one leg is the horizontal distance (12m) and the other leg is the vertical difference (5m). Step 4: Using Pythagoras theorem: rope length = √(12² + 5²) = √(144 + 25) = √169 = 13 metres.

Question 10

A person standing on the ground observes the top of a building at an angle of elevation of 45°. After walking 20 metres towards the building, the angle of elevation becomes 60°. Find the height of the building (in metres). [Use √3 ≈ 1.732]

Accepted Answer

Step 1: Let height of building = h, initial distance from building = x. From first position: tan(45°) = h/x, so h = x (since tan(45°) = 1). Step 2: From second position (20m closer): tan(60°) = h/(x−20), so h = (x−20)√3. Step 3: Equating the two expressions for h: x = (x−20)√3 → x = x√3 − 20√3 → x(√3−1) = 20√3 → x = 20√3/(√3−1). Step 4: 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. Step 5: h = x = 30 + 10√3 ≈ 30 + 17.32 = 47.32m.

Question 11

From a point on the ground 30 metres away from the base of a tower, the angle of elevation to the top is 30°. From the same point, the angle of elevation to a window on the tower is 15°. Find the vertical distance between the top of the tower and the window (in metres). [Use √3 ≈ 1.732, tan(15°) = 2 − √3]

Accepted Answer

Step 1: Horizontal distance from tower base = 30m. Step 2: Height to top of tower: tan(30°) = h_top/30, so h_top = 30 × tan(30°) = 30 × (1/√3) = 30/√3 = 10√3m. Step 3: Height to window: tan(15°) = h_window/30, so h_window = 30 × tan(15°) = 30 × (2−√3)m. Step 4: Vertical distance = h_top − h_window = 10√3 − 30(2−√3) = 10√3 − 60 + 30√3 = 40√3 − 60 = 20(2√3 − 3). Step 5: Numerically: 40√3 ≈ 69.28, so 69.28 − 60 = 9.28m ≈ 20(2×1.732−3) = 20(3.464−3) = 20(0.464) ≈ 9.28m.

Question 12

A man standing at point A observes the top of a tower at an angle of elevation of 60°. He walks 40 metres towards the base of the tower and observes the angle of elevation to be 75°. Find the height of the tower (in metres). [Use √3 ≈ 1.732]

Accepted Answer

Step 1: Let height of tower = h, initial distance from tower = x. From first position: tan(60°) = h/x, so h = x√3. Step 2: From second position (40m closer): tan(75°) = h/(x−40). Step 3: tan(75°) = tan(45°+30°) = (1+1/√3)/(1−1/√3) = (√3+1)/(√3−1) = 2+√3 ≈ 3.732. Step 4: Substitute h = x√3 into tan(75°) equation: 3.732 = x√3/(x−40). Step 5: 3.732(x−40) = 1.732x → 3.732x − 149.28 = 1.732x → 2x = 149.28 → x = 74.64m. Step 6: h = 74.64 × 1.732 ≈ 129.2m. Rounding to nearest clean value: h ≈ 130m (or 40(2+√3) ≈ 129.28m).

Question 13

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

Accepted Answer

Step 1: The ladder forms the hypotenuse of a right triangle with the wall and ground. Step 2: Angle between ladder and ground = 60°. Step 3: The distance from the wall (base) is the side adjacent to the 60° angle. Step 4: cos(60°) = adjacent/hypotenuse = distance/8. Step 5: cos(60°) = 1/2, so distance = 8 × (1/2) = 4 metres.

SBI Clerk Heights & Distances

Heights & Distances— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Heights & Distances — Practice Questions

60-Second Revision — Heights & Distances