Heights and Distances
Heights and Distances
Every heights-and-distances question is a right triangle with a tan ratio: height opposite the angle, horizontal distance next to it. NTPC asks a single angle of elevation, two angles from points on the same line, angles of depression from a tower or cliff, and objects on opposite sides.
1. Rules Box
tan(angle of elevation) = height/horizontal distance
Angle of depression from the top = angle of elevation from the bottom (alternate angles)
Two observation points d apart on the same side, angles α < β: h = d/(cot α − cot β)
Points on opposite sides of the tower: total distance = h(cot α + cot β)
| Angle | tan | Height for distance 1 | Distance for height 1 |
|---|---|---|---|
| 30° | 1/√3 | 1/√3 | √3 |
| 45° | 1 | 1 | 1 |
| 60° | √3 | √3 | 1/√3 |
| Useful values | Approximation |
|---|---|
| √3 | 1.732 |
| √2 | 1.414 |
| √3 + 1 | 2.732 |
| √3 − 1 | 0.732 |
✗ Angle of depression is measured from the vertical | ✓ It is measured below the horizontal line through the eye
✗ Elevation 30° then 60° from points 20 m apart, so height = 20 tan 60° | ✓ Use both angles: h = 20/(√3 − 1/√3) = 10√3 m
हिंदी नोट: उन्नयन कोण में tan = ऊँचाई/क्षैतिज दूरी। ऊपर से देखा गया अवनमन कोण नीचे से देखे गए उन्नयन कोण के बराबर होता है। दो बिंदुओं वाले प्रश्न में दोनों त्रिभुजों को ऊँचाई h से जोड़िए।
Exam Pointer: Verified NTPC pattern: a tower's height from a given distance and a single angle of elevation (March 2026 Graduate CBT-1). Two-angle, depression and opposite-side questions had no verified NTPC shift in our check and are tagged syllabus-based.
Pariksha Pattern: Every Way NTPC Asks This Topic
Pattern 1: Single angle of elevation
[PYQ: NTPC Graduate CBT-1 16-Mar-2026 Shift-1]
EXAM LEVEL
Q. From a point 30 m from the foot of a tower, the angle of elevation of its top is 60°. Find the height of the tower.
h = 30 tan 60° = 30√3 ≈ 51.96 m.
Answer: 30√3 m (about 51.96 m)
EXAMATLAS LEVEL
Q. A vertical pole breaks in a storm and its top touches the ground 12 m from the foot, making an angle of 30° with the ground. Find the original height of the pole.
The standing part = 12 tan 30° = 4√3 m. The broken part is the hypotenuse = 12/cos 30° = 24/√3 = 8√3 m. Original height = 12√3 ≈ 20.78 m.
Answer: 12√3 m (about 20.78 m)
Pattern 2: Two angles from points on the same side
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. The angles of elevation of the top of a tower from two points 20 m apart, on the same side and in line with its foot, are 30° and 60°. Find the height.
h = 20/(cot 30° − cot 60°) = 20/(√3 − 1/√3) = 20√3/2 = 10√3 ≈ 17.32 m.
Answer: 10√3 m
EXAMATLAS LEVEL
Q. A man sees the top of a tower at 30°. After walking 20 m towards it, he sees it at 45°. Find the height of the tower and his final distance from it.
h = 20/(√3 − 1) = 10(√3 + 1) ≈ 27.32 m. At 45° his distance equals the height, about 27.32 m.
Answer: 10(√3 + 1) m, about 27.32 m; the same distance
Pattern 3: Angles of depression from a height
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. From the top of a 75 m lighthouse the angle of depression of a ship is 30°. How far is the ship from the lighthouse?
Distance = 75 cot 30° = 75√3 ≈ 129.9 m.
Answer: 75√3 m
EXAMATLAS LEVEL
Q. From the top of a 60 m building, the angles of depression of the top and the bottom of a tower are 30° and 60°. Find the height of the tower.
Horizontal distance = 60 cot 60° = 20√3 m. The drop from the building top to the tower top = 20√3 × tan 30° = 20 m. Tower = 60 − 20 = 40 m.
Answer: 40 m
Pattern 4: Objects on opposite sides
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. Two men on opposite sides of a 30 m tower see its top at 30° and 60°. Find the distance between them.
30 cot 30° + 30 cot 60° = 30√3 + 10√3 = 40√3 ≈ 69.28 m.
Answer: 40√3 m
EXAMATLAS LEVEL
Q. From a point on a bridge 30 m above a river, the angles of depression of the two banks on opposite sides are 30° and 45°. Find the width of the river.
Width = 30 cot 30° + 30 cot 45° = 30√3 + 30 = 30(√3 + 1) ≈ 81.96 m.
Answer: 30(√3 + 1) m
60-Second Revision
- tan = height/distance; depression equals the matching elevation.
- Same side: h = d/(cot α − cot β).
- Opposite sides: distance = h(cot α + cot β).
- Keep √3 = 1.732 ready for numerical options.
Next Step: Trigonometry done. Solve the trigonometry set in the ExamAtlas RRB NTPC 2026 mock tests; put θ = 45° or 30° into any identity question to check the answer in seconds.