SSC Foundation Maths Height & Distance Formula, Concepts, Short Tricks

Height and Distance is one of the most important and scoring topics for competitive exams like SSC CGL, CHSL, MTS, and others. This chapter is rooted in basic trigonometry and focuses on real-life applications involving angles of elevation and depression, shadows, towers, trees, rivers, aeroplanes, and mountains. 

A strong grasp of standard trigonometric ratios (30°, 45°, 60°), triangle properties, and shortcut-based approaches such as the unitary method and complementary angle formulas can drastically reduce calculation time and improve accuracy. 

In this comprehensive guide, we cover Height & Distance formulas, core concepts, and short tricks through a wide range of exam-oriented problems from broken trees and changing shadows to swimmers crossing rivers and aeroplanes in motion, helping aspirants build both conceptual clarity and problem-solving speed for SSC examinations.

Problem 1: The Broken Tree

Problem Statement: A straight tree breaks due to a storm. The broken part bends so that the top of the tree touches the ground, making an angle of 30° with it. The distance from the foot of the tree to the point where the top touches the ground is 10 meters. Find the total height of the tree before it broke.

Solution Breakdown:

1. Let the original tree's base be O, the break point A, and the top touch the ground at B.

2. Given distance OB = 10 m and ∠OBA = 30°. The total height = AO (standing part) + AB (broken part, now hypotenuse).

3. Using the 30-60-90 triangle ratio: Opposite (AO) = 1 unit, Adjacent (OB) = √3 units, Hypotenuse (AB) = 2 units.

4. Since √3 units = 10 meters, then 1 unit = 10 / √3 meters.

5. Total height in units = 1 unit + 2 units = 3 units.

6. Total Height = 3 × (10 / √3) = 30 / √3.

7. Rationalizing: (30√3) / 3 = 10√3 meters.

Problem 2: Changing Angle of Elevation (30° to 45°)

Problem Statement: The angle of elevation of the sun changes from 30° to 45°. As a result, the length of a pillar's shadow decreases by 20 meters. Find the height of the pillar.

Solution Breakdown:

1. Let AB be the pillar's height. Initial shadow BC (angle 30°), final shadow BD (angle 45°).

2. Shadow decrease CD = 20 meters.

3. Unitary Method:

• For 45° (ΔABD): tan 45° = 1. If Height AB = 1 unit, Base BD = 1 unit.

• For 30° (ΔABC): tan 30° = 1/√3. If Height AB = 1 unit, Base BC = √3 units. (Height is already balanced).

4. Difference in shadow length BC - BD = (√3 - 1) units.

5. This difference equals 20 meters. So, (√3 - 1) units = 20 meters.

6. Pillar Height (1 unit) = 20 / (√3 - 1) meters.

7. Rationalizing: [20(√3 + 1)] / [(√3 - 1)(√3 + 1)] = [20(√3 + 1)] / (3 - 1) = 10(√3 + 1) meters.

Problem 3: Changing Angle of Elevation (60° to 30°)

Problem Statement: The shadow of a tower on a level plane is found to be 50 meters longer when the sun's elevation is 30° than when it is 60°. What is the height of the tower?

Solution Breakdown:

1. Let AB be the tower's height. Shadow BC for 60°, shadow BD for 30°.

2. Shadow increase CD = 50 meters.

3. Unitary Method:

• For 60° (ΔABC): tan 60° = √3. If AB = √3 units, then BC = 1 unit.

• For 30° (ΔABD): tan 30° = 1/√3. If AB = 1 unit, then BD = √3 units.

4. To balance height (AB), multiply 30° ratios by √3:

• New AB = 1 × √3 = √3 units.

• New BD = √3 × √3 = 3 units.

5. Height AB is now √3 units for both.

6. Difference in shadow length BD - BC = (3 - 1) = 2 units.

7. Given difference = 50 meters. So, 2 units = 50 meters, meaning 1 unit = 25 meters.

8. Height of tower (√3 units) = √3 × 25 = 25√3 meters.

Problem 4: Poles with Complementary Angles of Elevation

Problem Statement: The distance between two vertical pillars of lengths 16 m and 9 m is x meters. The angles of elevation of their respective tops from the bottom of the other are complementary to each other. Find the distance x.

Solution Breakdown:

1. Pillar AB = 16 m, Pillar CD = 9 m. Distance between bases BC = x.

2. Let ∠ACB = θ, then ∠DBC = (90° - θ).

3. In ΔABC: tan(θ) = 16 / x.

4. In ΔDCB: tan(90° - θ) = 9 / x, which is cot(θ) = 9 / x.

5. Multiply the equations: tan(θ) × cot(θ) = (16 / x) × (9 / x).

6. Since tan(θ) × cot(θ) = 1, then 1 = 144 / x².

7. x² = 144, so x = 12 meters.
   For any problem where two vertical poles of heights h₁ and h₂ are separated by a distance d, and the angles of elevation of their tops from the base of the other are complementary, the distance is given by the formula: d = √(h₁ × h₂) . In this case: x = √(16 × 9) = 12 meters.

Problem 5: Swimmer Crossing a River

Problem Statement: The two banks of a river are parallel. A swimmer starts from a point on one of the banks and swims in a straight line inclined to the bank at 45°. He reaches the opposite bank after swimming 20 meters. What is the width of the river?

Solution Breakdown:

1. The swimmer's path is the hypotenuse (20 m) of a right-angled triangle.

2. The angle with the starting bank is 45°. The width of the river is the side opposite this angle.

3. Using sin(45°) = Opposite / Hypotenuse:

• sin(45°) = Width / 20.

• 1/√2 = Width / 20.

• Width = 20 / √2.

4. Rationalizing: Width = (20√2) / 2 = 10√2 meters.

5. Approximate width: 10 × 1.414 = 14.14 meters.

Problem 6: Two Temples and Angles of Depression

Problem Statement: There are two temples, one on each bank of a river, opposite each other. One temple is 54 meters high. From the top of this temple, the angles of depression of the top and the foot of the other temple are 30° and 60° respectively. Find the height of the other temple.

Solution Breakdown:

1. Let the taller temple be AC (54 m), and the shorter be DE.

2. Angles of depression from A are 30° (to D) and 60° (to E). By alternate interior angles, ∠ADB' = 30° and ∠AEC = 60° (where B' is on DE).

3. Unitary Method: (Let CE = DB be the base)

• For 60° (ΔAEC): tan 60° = √3. If Height AC = √3 units, then Base CE = 1 unit.

• For 30° (ΔABD, where AB is horizontal): tan 30° = 1/√3. If Height AB = 1 unit, then Base DB = √3 units.

4. To make base equal, multiply 60° ratios by √3:

• New AC = √3 × √3 = 3 units.

• New CE = 1 × √3 = √3 units.

5. Now, AC = 3 units and AB = 1 unit.

6. Height of shorter temple (DE) = BC = AC - AB = 3 units - 1 unit = 2 units.

7. Given AC = 54 meters. So, 3 units = 54 meters.

8. 1 unit = 18 meters.

9. Height of shorter temple (2 units) = 2 × 18 = 36 meters.

Problem 7: Two Equal Poles on a Road

Problem Statement: Two poles of equal height are standing opposite each other on either side of a road, which is 100 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 30° and 60°. Find the height of the poles.

Solution Breakdown:

1. Poles AB and CD are of equal height. Road width BD = 100 m. Let P be the point between poles.

2. Angles of elevation are ∠APB = 60° and ∠CPD = 30°.

3. Unitary Method:

• For 60° (Pole AB): tan 60° = √3. If Base BP = 1 unit, Height AB = √3 units.

• For 30° (Pole CD): tan 30° = 1/√3. If Base DP = √3 units, Height CD = 1 unit.

4. To make pole heights equal, multiply 30° ratios by √3:

• New CD = 1 × √3 = √3 units.

• New DP = √3 × √3 = 3 units.

5. Both poles are now √3 units high.

6. Total road width BP + DP = 1 unit + 3 units = 4 units.

7. Given total width = 100 meters. So, 4 units = 100 meters, implying 1 unit = 25 meters.

8. Height of poles (√3 units) = √3 × 25 = 25√3 meters.

Problem 8: The Double Angle Problem (θ and 2θ)

Problem Statement: From a point 160 meters away from the foot of a tower, the angle of elevation is θ. On advancing 100 meters towards the tower, the angle of elevation becomes 2θ. Find the height of the tower.

Solution Breakdown:

1. Tower AB. Point D is 160 m from base B (DB = 160 m). Angle at D = θ.

2. Advance 100 m to point C (DC = 100 m). Angle at C = 2θ. Remaining base CB = 160 - 100 = 60 m.

3. Geometrical Shortcut (Exterior Angle Theorem): In ΔADC, exterior angle ∠ACB = 2θ.

4. ∠CAD + ∠ADC = 2θ. Since ∠ADC = θ, then ∠CAD = θ.

5. Thus, ΔADC is isosceles with ∠CAD = ∠ADC = θ.

6. Therefore, AC = DC = 100 meters.

7. In right-angled ΔABC: Hypotenuse AC = 100 m, Base CB = 60 m.

8. Using Pythagorean theorem (or 6-8-10 triplet scaled by 10):

• AB² = AC² - CB² = 100² - 60² = 10000 - 3600 = 6400.

• AB = √6400 = 80 meters.

9. Height of tower AB = 80 meters.

Problem 9: Complementary Angles in a Straight Line

Problem Statement: The angles of elevation of the top of a tower from two points at distances of 10 m and 5 m from the base of the tower and in the same straight line with it are complementary. Find the height of the tower.

Solution Application:

When angles of elevation from two points (d₁ and d₂) on the same straight line from the base of a tower are complementary, the height (h) of the tower is given by: h = √(d₁ × d₂) .

Calculation:

• d₁ = 10 m

• d₂ = 5 m

• h = √(10 × 5) = √50 = 5√2 meters.

Problem 10: Calculating Airplane Speed

Problem Statement: The angle of elevation of an airplane from a point on the ground is 60°. After a flight of 10 seconds, the elevation changes to 30°. If the airplane is flying at a constant height of 1500√3 meters, find the speed of the plane.

Solution Breakdown:

1. Speed = Distance / Time. Time = 10 seconds. We need horizontal distance traveled.

2. Let the plane's height be AB = CD = 1500√3 m. Observer at O.

3. Initial angle ∠AOB = 60°. Final angle ∠COD = 30°.

4. Calculate Horizontal Distances:

• For 60° (ΔAOB): tan 60° = AB / OB => √3 = 1500√3 / OB => OB = 1500 meters.

• For 30° (ΔCOD): tan 30° = CD / OD => 1/√3 = 1500√3 / OD => OD = 1500√3 × √3 = 4500 meters.

5. Distance traveled (horizontal) = OD - OB = 4500 - 1500 = 3000 meters.

6. Speed = 3000 meters / 10 seconds = 300 m/s.

Problem 11: The Mountain Climbing Problem

Problem Statement: At the foot of a mountain, the elevation of its summit is 45°. After ascending 3 km up the mountain on a slope of 30° inclination, the elevation of the summit is found to be 60°. Find the height of the mountain.

Solution Application:

For this specific scenario (initial angle 45°, slope 30°, final angle 60°), the height of the mountain (H) can be calculated directly using the distance ascended (d): Height (H) = (√3 + 1) / 2 × d.

Calculation:

• Distance ascended (d) = 3 km.

• Height = (√3 + 1) / 2 × 3 = 3(√3 + 1) / 2 km.