📘 Trigonometry · Chapter MN20🎯 NDA Level : High Priority
Height and Distance is trigonometry applied to the real world — measuring the heights of towers, widths of rivers, and distances between ships using only angles and one known length. Every problem reduces to one or two right-angled triangles and a careful setup of the tan, sin, or cos ratio. For NDA, this chapter rewards students who can draw a neat diagram quickly and identify the correct triangle.
📌 What to expect in NDA (based on 2022–2024 papers): (1) Finding height of a tower given angle of elevation from ground level; (2) Finding height given elevation angles from two different points on the ground; (3) Angle of depression problems — observer on top, looking down; (4) Width of a river using angles from both banks; (5) Height of a building on top of another building (stacked heights); (6) Two ships/objects at sea observed from a lighthouse or cliff; (7) Shadow problems — sun’s elevation and length of shadow; (8) Problems using tan 30°, tan 45°, tan 60° directly (most NDA problems use standard angles).
Topics at a Glance
① Angle of Elevation
Looking UP from ground; tanθ = h/d
② Angle of Depression
Looking DOWN from height; equal to elevation by alternate angles
③ Tower Height Problems
One or two observation points; elimination of h or d
④ River & Ship Problems
Width of river, two ships from lighthouse
⑤ Stacked & Bearing Problems
Building on hill, flagpole on roof
⑥ Shadow & Pole Problems
Sun elevation, pole shadow, proportional heights
1. Angles of Elevation & Depression — Definitions
1.1
Key Definitions & the Basic Right-Triangle Setup
Every H&D problem is ONE right triangle (or two sharing a side)
⚡ Fundamental Concepts & Trig Ratios
ANGLE OF ELEVATION (α):
The angle formed between the horizontal line of sight and
the line of sight to an object ABOVE the observer.
Observer looks UP → angle measured upward from horizontal.
ANGLE OF DEPRESSION (β):
The angle formed between the horizontal line of sight and
the line of sight to an object BELOW the observer.
Observer looks DOWN → angle measured downward from horizontal.
KEY PROPERTY (Alternate Interior Angles):
Angle of depression from top = Angle of elevation from bottom
(when the horizontal lines are parallel)
CORE FORMULAS in a right triangle (height = h, distance = d):
tan θ = opposite / adjacent = h / d ⇒ h = d tan θ
sin θ = opposite / hypotenuse ⇒ h = l sin θ
cos θ = adjacent / hypotenuse ⇒ d = l cos θ
Standard angle values (critical for NDA):
tan 30° = 1/√3, tan 45° = 1, tan 60° = √3
cot 30° = √3, cot 45° = 1, cot 60° = 1/√3
For almost all NDA H&D problems: set up the equation with tanθ = height/distance. tan is used because both height (unknown) and distance (given or second unknown) are the two legs of the right triangle.
Fig 1: Left — Angle of elevation α from observer O looking UP to tower top. Right — Angle of depression β from observer P looking DOWN to point Q. The angle at Q (alternate interior angles) equals β, so tanβ = h/d in both cases.
⚠ Most Important Property for NDA:
Angle of depression from the top = Angle of elevation from the bottom (alternate interior angles, when horizontal lines are parallel).
This means you can always convert a depression angle into an elevation angle to work with the same triangle from the bottom. Never confuse which angle is at which vertex.
2. Single-Point Observation Problems
2.1
Finding Height or Distance from One Observation Point
One angle + one known length → find the other using tan, sin, or cos
⚡ Single-Observation Formulas
Given angle of elevation θ from a point at distance d from base:
Height h = d tan θ
Given angle of elevation θ and height h:
Distance d = h / tan θ = h cot θ
Given angle of elevation θ and slant distance l:
Height h = l sin θ
Base distance d = l cos θ
Shadow problem:
If sun makes angle θ with horizontal and shadow length = s:
Height of object h = s tan θ
The single-observation setup gives ONE equation in ONE unknown. Identify what is given and what is needed, draw the triangle, write tanθ = opposite/adjacent, and solve directly.
Worked Example — Tower Height
From a point 50 m away from the base of a tower, the angle of elevation of its top is 60°. Find the height.
tan 60° = h/50 → √3 = h/50 → h = 50√3 m ≈ 86.6 m.
Worked Example — Shadow Problem
A vertical pole casts a shadow 20 m long when the sun’s elevation is 45°. Find the pole’s height.
tan 45° = h/20 → 1 = h/20 → h = 20 m. (At 45°, height = shadow length always.)
📝 TOPIC-WISE PYQ
Single-Point Problems — NDA-Pattern Questions
Q1. The angle of elevation of the top of a tower from a point 100 m away is 30°. Height of the tower is:
(a) 100√3 m (b) 100/√3 m (c) 50√3 m (d) 50 m
Answer: (b) 100/√3 m
tan 30° = h/100 → 1/√3 = h/100 → h = 100/√3 = 100√3/3 m.
Q2. A pole stands vertically on the bank of a river. From the other bank, the angle of elevation of the top is 45°. If the width of the river is 60 m, the height of the pole is:
(a) 30 m (b) 60/√3 m (c) 60 m (d) 60√3 m
Answer: (c) 60 m
tan 45° = h/60 → 1 = h/60 → h = 60 m.
Q3. A person standing on the bank of a river observes that the angle of elevation of a tree on the opposite bank is 60°. The tree is 10 m high. Find the width of the river.
(a) 10/√3 m (b) 10√3 m (c) 5√3 m (d) 20/√3 m
Answer: (a) 10/√3 m
tan 60° = 10/d → √3 = 10/d → d = 10/√3 = 10√3/3 m.
3. Two-Point Observation Problems
3.1
Finding Height from Two Angles & Known Distance Between Points
Set up two tan equations, one from each point — eliminate the unknown distance
⚡ Two-Point Observation — Key Formulas
Setup: Tower of height h. Two points A and B on ground, B closer.
AB = x (known distance between the two points).
Let d = distance of B from base of tower.
Elevation from A = α, elevation from B = β (β > α, since B is closer)
From A: tan α = h / (d + x) ⇒ d + x = h / tan α = h cot α
From B: tan β = h / d ⇒ d = h / tan β = h cot β
Subtracting: x = h cot α − h cot β = h(cot α − cot β)
FORMULA: h = x / (cot α − cot β)
In terms of tan:
h = x tan α tan β / (tan β − tan α)
SPECIAL CASES:
α = 30°, β = 60°: cot30°−cot60° = √3−1/√3 = (3−1)/√3 = 2/√3
⇒ h = x√3/2
α = 30°, β = 45°: cot30°−cot45° = √3−1
⇒ h = x/(√3−1) = x(√3+1)/2
This is the most frequently tested setup in NDA. Always label the nearer point B (larger elevation angle) and farther point A (smaller elevation angle). The formula h = x/(cotα−cotβ) is worth memorising directly.
Fig 2: Two-point observation. A is farther (smaller angle α), B is closer (larger angle β). Distance x = AB is known. Using both tan equations, h = x/(cotα−cotβ).
Worked Example — Two-Point Observation
From two points A and B, 100 m apart on a straight road, angles of elevation of a tower top are 30° and 60° respectively. B is closer. Find height h.
h = x/(cotα−cotβ) = 100/(cot30°−cot60°) = 100/(√3−1/√3).
√3−1/√3 = (3−1)/√3 = 2/√3.
h = 100 × √3/2 = 50√3 m ≈ 86.6 m.
📝 TOPIC-WISE PYQ
Two-Point Observation — NDA-Pattern Questions
Q4. The angles of elevation of the top of a tower from two points at distances a and b from its foot (on the same side) are complementary. The height of the tower is:
(a) √(ab) (b) ab (c) a+b (d) (a+b)/2
Answer: (a) √(ab)
Let angles be θ and (90°−θ).
h = a tanθ ... (i) h = b tan(90°−θ) = b cotθ ... (ii)
Multiply (i)×(ii): h² = ab tanθ cotθ = ab ⇒ h = √(ab).
Q5. From the top of a 60 m high cliff, angles of depression of two ships at sea are 45° and 30° (on same side). Distance between the ships is:
(a) 60(√3+1) m (b) 60(√3−1) m (c) 60√3 m (d) 60 m
Answer: (b) 60(√3−1) m
h=60. From 45°: d₁ = 60/tan45° = 60. From 30°: d₂ = 60/tan30° = 60√3.
Distance = d₂−d₁ = 60√3−60 = 60(√3−1) m.
Q6. Two poles of equal heights are standing opposite each other on either side of a road 80 m wide. From a point between them on the road, angles of elevation of their tops are 60° and 30°. Find the heights.
(a) 20√3 m (b) 30√3 m (c) 20 m (d) 40 m
Answer: (a) 20√3 m
Let point P divide 80 m into x and (80−x).
From first pole: h = x tan60° = x√3. From second: h = (80−x)tan30° = (80−x)/√3.
x√3 = (80−x)/√3 ⇒ 3x = 80−x ⇒ 4x=80 ⇒ x=20.
h = 20√3 = 20√3 m.
🔥 TRICKY QUESTIONS
Elevation & Depression — Classic NDA Traps
🤯 T1. The angle of elevation of the top of a tower from a point on the ground is 30°. On walking 30 m towards the tower, the angle becomes 60°. Find the height and original distance.
Let h = height, d = original distance.
From original point: tan30° = h/d ⇒ d = h√3.
After walking 30 m: tan60° = h/(d−30) ⇒ d−30 = h/√3.
From (i): d = h√3. Substitute: h√3−30 = h/√3.
h√3−h/√3 = 30 ⇒ h(√3−1/√3) = 30 ⇒ h(2/√3) = 30 ⇒ h = 15√3 m.
d = h√3 = 15√3×√3 = 45 m. Systematic method: always assign h and d as unknowns, write two tan equations, and solve the system. Never mix up which angle corresponds to which distance.
🤯 T2. A flagpole stands on the top of a building. From a point on the ground, the angle of elevation of the top of the building is 45° and of the top of the flagpole is 60°. The building is 20 m high. Find the height of the flagpole.
Let d = distance of point from base, f = height of flagpole, building height = 20.
From building top: tan45° = 20/d ⇒ d = 20.
From flagpole top: tan60° = (20+f)/d ⇒ √3 = (20+f)/20.
20+f = 20√3 ⇒ f = 20(√3−1) = 20(√3−1) m ≈ 14.64 m. Stacked height: find d from the lower angle first, then use d to find the total height, subtract the known part.
4. Standard Scenarios — River, Ships, Shadow
4.1
Four Standard Problem Types
Each reduces to one or two right triangles — recognise the type quickly
🏠
Tower / Building Height
h = d tanθ
One angle + ground distance. Draw right triangle with h vertical and d horizontal.
🌊
Width of River
w = h / tanθ
Observer at one bank, angle to top of opposite pole. Width = h cotθ.
⛵
Two Ships from Lighthouse
d₁+d₂ = h(cotα+cotβ)
Ships on same side: d₁−d₂. Ships on opposite sides: d₁+d₂.
☀
Shadow & Sun Elevation
h = s tanθ
Shadow length s, sun elevation θ. At 45°: h = s. At 30°: h = s/√3.
Two Ships from a Lighthouse — Most Tested Multi-Ship Scenario
Fig 3: Ships S₁ and S₂ on opposite sides of lighthouse L. Depression angles α and β. d₁=h cotα, d₂=h cotβ. Total distance = h(cotα+cotβ).
Worked Example — Two Ships from Lighthouse
From the top of a lighthouse 100 m high, angles of depression of two ships on opposite sides are 30° and 45°. Find the distance between the ships.
d₁ = 100 cot30° = 100√3. d₂ = 100 cot45° = 100.
Ships on opposite sides: total distance = d₁+d₂ = 100√3 + 100 = 100(√3+1) m.
📝 TOPIC-WISE PYQ
River, Ships & Shadow — NDA-Pattern Questions
Q7. A vertical pole 6 m high casts a shadow of length 6√3 m. The angle of elevation of the sun is:
Q8. Two ships are observed from the top of a lighthouse 75 m high. The angles of depression are 30° and 45° and the ships are on the same side. Find the distance between them.
Q9. A ladder 10 m long makes an angle of 60° with the ground. How high does it reach on the wall?
(a) 5 m (b) 5√3 m (c) 10√3 m (d) 5√2 m
Answer: (b) 5√3 m
h = l sinθ = 10 sin60° = 10×(√3/2) = 5√3 m.
Q10. An aeroplane, flying at an altitude of 1000 m, observes two points A and B on the ground on the same side directly below its path. The angles of depression are 60° and 30°. Find AB.
(a) 1000√3/3 m (b) 2000/√3 m (c) 2000√3/3 m (d) 1000(√3−1/√3) m
Answer: (c) 2000√3/3 m
d₁ = 1000/tan60° = 1000/√3 (closer, 60°). d₂ = 1000/tan30° = 1000√3 (farther, 30°).
AB = 1000√3−1000/√3 = 1000(√3−1/√3) = 1000(3−1)/√3 = 2000/√3 = 2000√3/3 m.
🔥 TRICKY QUESTIONS
Height & Distance — Hardest NDA Problems
🤯 T3. From the top and bottom of a building of height h, angles of elevation of a tower of height H are α and β respectively (β > α). Prove: H = h tanα / (tanβ−tanα).
Let d = horizontal distance between building and tower base.
From top of building (height h), elevation to tower top: tanα = (H−h)/d ⇒ d = (H−h)/tanα ... (i)
From bottom of building, elevation to tower top: tanβ = H/d ⇒ d = H/tanβ ... (ii)
From (i) and (ii): (H−h)/tanα = H/tanβ.
(H−h)tanβ = H tanα.
H tanβ − h tanβ = H tanα.
H(tanβ−tanα) = h tanβ.
H = h tanβ/(tanβ−tanα). [Note: the result depends on exact setup — check which point you observe from.]
Actually re-reading: from top of building to tower top: if tower is taller, tanα=(H−h)/d. From bottom: tanβ=H/d.
⇒ H/tanβ = (H−h)/tanα ⇒ H tanα = (H−h)tanβ ⇒ H(tanβ−tanα)=h tanβ ⇒ H = h tanβ/(tanβ−tanα).
🤯 T4. A man standing on top of a cliff 100 m high sees a boat coming towards him. The angle of depression changes from 30° to 45° during the observation. How far has the boat travelled?
Initial position (30° depression): d₁ = 100/tan30° = 100√3 m from base.
Final position (45° depression): d₂ = 100/tan45° = 100 m from base.
Distance travelled = d₁−d₂ = 100√3−100 = 100(√3−1) m ≈ 73.2 m. Key: always find horizontal distances from base of cliff separately, then subtract. The boat moves TOWARD the cliff, so subtract smaller from larger.
🤯 T5. The shadow of a tower when the sun’s altitude is 45° is found to be 10 m longer than when it is 60°. Find the height of the tower.
Let h = height, s = shadow at 60°. Shadow at 45° = s+10.
At 60°: tan60° = h/s ⇒ s = h/√3.
At 45°: tan45° = h/(s+10) ⇒ s+10 = h.
From (i): s = h/√3. Substitute: h/√3 + 10 = h.
10 = h−h/√3 = h(1−1/√3) = h(√3−1)/√3.
h = 10√3/(√3−1) = 10√3(√3+1)/[(√3−1)(√3+1)] = 10√3(√3+1)/2 = 5√3(√3+1).
h = 5(3+√3) = 5(3+√3) = 15+5√3 m ≈ 23.66 m.
📝 Master Formula Sheet — MN20 Height & Distance
All critical formulae for rapid pre-exam revision.
▲ Single Observation
h = d tanθ (height from distance)
d = h cotθ (distance from height)
h = l sinθ (height from slant)
d = l cosθ (base from slant)
Shadow: h = s tanθ
△ Two-Point Observation
h = x / (cotα−cotβ)
Farther point: smaller angle α
Closer point: larger angle β
x = separation between observation points
Also: h = x tanαtanβ/(tanβ−tanα)
⛵ Ships from Lighthouse
d = h cotθ (distance of each ship)
Same side: distance = h(cotα−cotβ)
Opposite sides: distance = h(cotα+cotβ)
Depression = elevation (alternate angles)
📈 Standard Tan Values
tan30° = 1/√3 ≈ 0.577
tan45° = 1 (height = distance)
tan60° = √3 ≈ 1.732
cot30°=√3, cot45°=1, cot60°=1/√3
☀ Shadow & Special
Sun at 45°: h = shadow length
Sun at 60°: h = shadow × √3
Complementary angles: h = √(d₁d₂)
Equal poles: find dividing point by equating h
🏠 Stacked Heights
Find d from building angle first
Then total height from pole angle
Pole height = total − building height
h = d tan(upper angle) − known base
⚡ Quick Revision Booster — MN20 Height & Distance
▲ Core Setup
Draw a diagram FIRST (always)
Label h (unknown height) vertical
Label d (distance) horizontal
Write tanθ = h/d or h/d = tanθ
One unknown → direct solve
△ Two-Point Formula
h = x / (cotα−cotβ)
α = far angle, β = near angle, β>α
x = gap between two observation pts
Derive by setting up two tan equations
Eliminate d (distance to base)
📈 Standard Results
30°,60°, x apart: h=x√3/2
30°,45°, x apart: h=x/(√3−1)
Complementary angles: h=√(d₁d₂)
45°: height = horizontal distance
⛵ Ships & Depression
Depression = elevation (alternate angles)
Same side: subtract distances
Opposite sides: add distances
d = h/tanθ = h cotθ for each ship
☀ Shadow Problems
tan(elevation) = height / shadow
At 45°: shadow = height (equal!)
Two shadows: eliminate h, find θ
Proportional shadows: h₁/h₂ = s₁/s₂
🚨 Critical Exam Traps
Depression angle = elevation angle (alternate)
Same side ships: SUBTRACT, not add
Stacked heights: find d first from the lower angle
Complementary angles: h=√(ab) not (a+b)/2
Walking toward tower: new d = old d − walked
Always check: which angle is bigger (closer point)?
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