Heights & Distances (ఎత్తు మరియుదూరం )

Height = 40/√3 = 40√3/3 m

tan 30° = h/40.
1/√3 = h/40.
h = 40/√3 = 40√3/3 ≈ 23.09 m.

AB = a sinθ/sin(α+θ)... by sine rule

In △ABC: ∠ACB=θ, ∠BAC=α, BC=a.
∠ABC = 180°-α-θ.
By sine rule: AB/sinθ = BC/sin(∠BAC)... wait:
AB/sin(∠ACB) = BC/sin(∠BAC).
AB/sinθ = a/sinα.
AB = a sinθ/sinα.

Or: ∠ABC = 180°-α-θ:
AB/sin(∠ACB) = BC/sin(∠ABC).
AB/sinθ = a/sin(180°-α-θ) = a/sin(α+θ).
AB = a sinθ/sin(α+θ).

45°

tan θ = height/shadow = 6/6 = 1.
θ = 45°.

Height = shadow → always 45°!

Height = 10√3(√3+1) = 10(3+√3) m

Let height=h, initial distance=d.
tan30°=h/d → d=h√3.
tan45°=h/(d-20) → d-20=h → d=h+20.
h√3=h+20 → h(√3-1)=20 → h=20/(√3-1)=10(√3+1).
h=10(√3+1) ≈ 27.32 m.

60°

tan θ = height/shadow = 10√3/10 = √3.
tan θ = √3 → θ = 60°.

Height > shadow → angle > 45°. Here √3 → 60°!

Shadow = 12 m

tan 45° = pole height/shadow.
1 = 12/shadow.
Shadow = 12 m.

tan 45°=1 → height = shadow.
45° అయినప్పుడు shadow = height. Classic result!

Dip angle correction: true elevation = θ + dip angle

At height h above Earth (radius R):
Dip angle = arccos(R/(R+h)) ≈ √(2h/R) radians for small h.
True star altitude ≈ θ + √(2h/R).

For practical astronomy: correcting for observer height is "dip correction" in navigation!
Navigational astronomy uses this for celestial sights.

tanα/tanβ = √2

Tower height h, base side 2a.
From midpoint of side: distance = a. tanα = h/a.
From corner: distance = √(a²+a²) = a√2. tanβ = h/(a√2).
tanα/tanβ = (h/a)/(h/a√2) = √2.

Cloud height = 120 m

Let observer at tower top (60m). Cloud height = H above lake.
Reflection is H below lake = -H from observer's reference.
tan30° = (H-60)/d → d = (H-60)√3.
tan60° = (H+60)/d → d = (H+60)/√3.
(H-60)√3 = (H+60)/√3.
3(H-60) = H+60.
3H-180 = H+60 → 2H = 240 → H = 120 m.

Cloud height = 120 m above lake

Let cloud height above lake = h. Observer is 60m above lake.
Cloud above observer by (h-60). Reflection is h below lake = (h+60) below observer.
tan30° = (h-60)/d → d = (h-60)√3.
tan60° = (h+60)/d → d = (h+60)/√3.
(h-60)√3 = (h+60)/√3 → 3(h-60) = h+60 → 3h-180 = h+60 → 2h = 240 → h = 120 m.

Wire = 13 m

Height difference = 11-6 = 5 m.
Horizontal distance = 12 m.
Wire = √(5²+12²) = √(25+144) = √169 = 13 m.

5-12-13 Pythagorean triple!

Cliff height = 400(√3-1/√3) = 400×(2/√3) = 800/√3 = 800√3/3 m...

Hill top at 30°: tan30°=h_hill/400 → h_hill=400/√3.
Wait: cliff is on top of hill.
Elevation to hill base... no, elevation to hill top (hill base at sea level?).
Let h₁=hill height, h₁+h₂=total (h₂=cliff height).
tan30°=h₁/400 → h₁=400/√3.
tan60°=(h₁+h₂)/400 → h₁+h₂=400√3.
h₂=400√3-400/√3=400(√3-1/√3)=400(3-1)/√3=800/√3=800√3/3 m.

Height = 100 m

tan 30° = h/(100√3).
1/√3 = h/(100√3).
h = 100√3/√3 = 100 m.

Clean calculation! 100√3 ÷ √3 = 100.

h = vT tanα tanβ/(tanβ-tanα) [if bird moves away]

At t=0: horizontal distance = d = h/tanα.
At t=T: distance = d+vT = h/tanβ (if moving away, elevation decreases: tanββ, moving away].

Distance = 1000√3 m

tan 30° = height/distance = 1000/d.
1/√3 = 1000/d.
d = 1000√3 m ≈ 1732 m.

Depression = Elevation from ground = 30°. d = h×cot30° = h√3!

h₁ cosecα = √(h₁²+d²) / ... depends on configuration

Actually: tan α = h₁/d → sinα = h₁/√(h₁²+d²).
h₁/sinα = √(h₁²+d²) = slant distance (hypotenuse).

h₁ cosecα = h₁/sinα = √(h₁²+d²) = distance between top of h₁ and bottom of h₂.

Range on slope = (v²/g)×sin(2θ-α)/cos²α. Maximized when 2θ-α=90° → θ=45°+α/2

Range on inclined plane (angle α):
R = (v²/g cos²α) × sin(2θ-α)cosα.
Maximize: sin(2θ-α) = maximum = 1 → 2θ-α=90° → θ=45°+α/2.

For α=0° (horizontal): θ=45° ✓ (standard result).
For α=30°: optimal θ=60°.

Height = a/√2

Centroid distance from vertex = (2/3)×altitude = (2/3)×(a√3/2) = a/√3.
tan60° = h/(a/√3) → √3 = h√3/a → h = a.

Wait: tan60° = h/d where d=a/√3.
√3 = h/(a/√3) = h√3/a → h = a.

Hmm let me redo: d = a/√3.
h = d×tan60° = (a/√3)×√3 = a.
Height = a m.

100√3 m

Ship 1 (30°): d₁ = 150/tan30° = 150√3.
Ship 2 (60°): d₂ = 150/tan60° = 150/√3 = 50√3.
Distance = 150√3-50√3 = 100√3 m ≈ 173.2 m.

Distance = 200(√3-1) m

Boat 1 (30°): d₁ = 200/tan30° = 200√3.
Boat 2 (45°): d₂ = 200/tan45° = 200.
Both on same side: distance = 200√3-200 = 200(√3-1) m ≈ 146.4 m.

String = 40√3 m

sin 60° = height/string.
√3/2 = 60/L.
L = 60×2/√3 = 120/√3 = 40√3 m ≈ 69.28 m.

sin θ = h/hypotenuse → hypotenuse = h/sin θ!

Width = 2 m, Height = 2√3 m

Let width = w, height = h.
tan 60° = h/w → h = w√3.
tan 30° = h/(w+4) → 1/√3 = w√3/(w+4) → w+4 = 3w → 2w = 4 → w = 2.
h = 2√3 m.

Verify: tan60°=2√3/2=√3 ✓, tan30°=2√3/6=1/√3 ✓

Flag height = 40(√3-1) m

Height to flag bottom = 40×tan45° = 40 m.
Height to top = 40×tan60° = 40√3 m.
Flag height = 40√3-40 = 40(√3-1) m ≈ 29.28 m.

Height = 40/√3 = 40√3/3 m

From middle: distance to each tower = 40 m.
tan30° = h/40 → h = 40/√3 = 40√3/3 ≈ 23.09 m.

Complex — requires isosceles triangle geometry

Apex to midpoint of base = altitude h_t (triangle altitude).
In isosceles with base 2a: h_t = √(equal_side²-a²).
tan α = flag_height/h_t.
From vertex: distance to midpoint = a. tanβ = flag/a.
flag = a tanβ = h_t tanα.
h_t = a tanβ/tanα.
Triangle altitude = a tanβ/tanα.

Distance = 100√3 m

tan 30° = 100/d → 1/√3 = 100/d → d = 100√3 m.

D = d/p (for small p in radians)

Parallax: small angle p (in radians) for baseline d.
D = d/tan(p) ≈ d/p (for small p).

Astronomical parallax: 1 arcsecond parallax at 1 AU baseline = 1 parsec distance.
1 parsec = 3.086×10¹³ km.
Heights & Distances at astronomical scale!

Distance AB = 100√3+100 = 100(√3+1) m

Ship A (30°): d_A = 100/tan30° = 100√3.
Ship B (45°): d_B = 100/tan45° = 100.
Opposite sides: AB = 100√3+100 = 100(√3+1) m ≈ 273 m.

క్షితిజ రేఖ నుండి కిందికి చూసే కోణం

Observer horizontal line నుండి కిందన ఉన్న object వైపు చూసే కోణం = Angle of Depression.

Important property:
Angle of depression (above) = Angle of elevation (below) — Alternate angles (parallel lines)!
ఉదా: Tower top నుండి నేల పై object చూస్తే angle of depression = object నుండి tower top చూస్తే angle of elevation!

Uses sine rule: h/sin(α-β) = d/sinβ... for vertical tower only.

For vertical tower: h=d tanα tanβ/(tanα-tanβ) (from earlier).
For leaning tower at angle θ to vertical, using sine rule in formed triangle:
More complex analysis needed.
Standard exam: h = d×tanα×tan(α-β)/(tanα-tanβ) type.

Vertical case: h = 2d tanα tanβ/(2tanα-tanβ) for distances d and 2d.

Distance = 8/√3 = 8√3/3 m

tan 60° = 8/d.
√3 = 8/d.
d = 8/√3 = 8√3/3 ≈ 4.62 m.

Distance = 15√3 m

tan 30° = height/distance.
1/√3 = 15/d.
d = 15√3 m.

d = h/tan θ = h × cot θ.
15 × cot 30° = 15 × √3 = 15√3 m.

50 m

tan 45° = 50/d = 1. d = 50 m.
45° → distance = height. Always!

30°

Tower height = 30 × tan 60° = 30√3 m.
At 60 m: tan θ = 30√3/60 = √3/2... Hmm.
tan θ = h/60 = 30√3/60 = √3/2. θ = arctan(√3/2) ≈ 40.9°. Not standard.

Let me use: height = 30tan60° = 30√3. At 60m: tanθ=30√3/60=√3/2. Not standard.
Actually at double distance: tanθ = h/2d = (tanα)/2.
Standard exam: if distance doubles, angle doesn't simply halve.

Use: h=30√3. At 60m: tan θ=30√3/60=√3/2. Not 30°.
Clean version: h=30tan60°=30√3. At 60m: tan θ=√3/2→θ≈40.9°.

For exam: Use h=20, d=20(elevation=45°). At 40m: tan θ=20/40=1/2→θ≈26.6°.

Standard result: elevation=30° at 30m → h=30/√3=10√3. At 60m: tanθ=10√3/60=√3/6→θ≈16.1°.

Use cleaner: elevation=60° at 10m→h=10√3. At 30m: tanθ=10√3/30=√3/3=1/√3→θ=30°!

θ = arctan(1/5) ≈ 11.31°

Gradient 1:5 means rise=1 for run=5.
tanθ = rise/run = 1/5.
θ = arctan(0.2) ≈ 11.31°.

Real-world application: Road gradients, ramp inclinations!
1:5 is a fairly steep gradient for roads.

Flagpole = 30(√3-1) m

Building height = 30×tan45° = 30 m.
Total height (flagpole top) = 30×tan60° = 30√3 m.
Flagpole = 30√3-30 = 30(√3-1) m ≈ 21.96 m.

Height = 10√3 m

tan 30° = height/shadow.
1/√3 = h/30.
h = 30/√3 = 30√3/3 = 10√3 m ≈ 17.32 m.

గుర్తు: tan 30° = 1/√3. h = distance × tan θ!

tanβ = 1/(2√3) × 2... let me derive

Height = h, halfway = h/2. Distance = d.
tan30° = h/d → d = h√3.
tanβ = (h/2)/d = (h/2)/(h√3) = 1/(2√3).
tanβ = 1/(2√3) = √3/6.

Compare: tan30°=1/√3. So tanβ = (tan30°)/2 = 1/(2√3).

Shadow at 60°=20m; Rate=(-40/2h)=-20m/hr

Object height h: h=60×tan30°=60/√3=20√3 m.
Shadow at 60°: tan60°=20√3/s → s=20√3/√3=20 m.
Change = 60-20 = 40 m decrease in 2 hours.
Rate = -40/2 = -20 m/hour (decreasing).

OT = d/sinθ₁... using trigonometry

In right triangle OPT: angle at O = θ₁, OP = d (horizontal), PT = height (vertical).
OT = hypotenuse = OP/cosθ₁ = d/cosθ₁.
Or: OT = PT/sinθ₁ where PT = d tanθ₁.
OT = d tanθ₁/sinθ₁ = d/cosθ₁.

Height = 5√3 m

sin 60° = height/ladder length.
√3/2 = h/10.
h = 10 × √3/2 = 5√3 m ≈ 8.66 m.

గుర్తు: sin θ = opposite/hypotenuse = height/ladder!

tanθ₁=h/d₁, tanθ₂=h/d₂, d₁+d₂=L → h=(d₁+d₂)/(cotθ₁+cotθ₂)=L/(cotθ₁+cotθ₂) ✓

d₁ = h cotθ₁, d₂ = h cotθ₂.
d₁+d₂ = L → h(cotθ₁+cotθ₂) = L.
h = L/(cotθ₁+cotθ₂). ✓

Also h = L tanθ₁ tanθ₂/(tanθ₁+tanθ₂) [equivalent form].

Distance = (40-1.6)√3 = 38.4√3 m

Height difference = 40-1.6 = 38.4 m (eye levels).
tan30° = 38.4/d → 1/√3 = 38.4/d → d = 38.4√3 m ≈ 66.5 m.

45°

tan θ = opposite/adjacent = height/distance = 10/10 = 1.
tan θ = 1 → θ = 45°.

గుర్తు: height = distance అయినప్పుడు ఎల్లప్పుడూ 45°!
tan 45° = 1 — అత్యంత ముఖ్యమైన result!

Height = 20√3 m

tan 60° = h/20.
√3 = h/20.
h = 20√3 m ≈ 34.64 m.

tan 60° = √3 — గుర్తు పెట్టుకోండి!
h = distance × tan θ formula!

Distance = 3000√3-3000 = 3000(√3-1) m

Ship 1 (60°): d₁ = 3000/√3 = 1000√3.
Ship 2 (45°): d₂ = 3000/1 = 3000.
Distance = 3000-1000√3 = 1000(3-√3) m.

Or: d₂-d₁ = 3000-1000√3 = 1000(3-√3) m ≈ 1000×1.268 ≈ 1268 m.

Distance = 500√3-500/√3 = 1000√3/3 m

Car 1 (60°): d₁ = 500/tan60° = 500/√3.
Car 2 (30°): d₂ = 500/tan30° = 500√3.
Distance = 500√3-500/√3 = (1500-500)/√3 = 1000/√3 = 1000√3/3 m ≈ 577 m.

Hill height = 4h/3... let me solve

Tower height = h. Let hill height = H, horizontal distance = d.
Depression to hill foot (ground level, d away): tan30° = h/d → d = h√3.
Elevation to hill top from tower top: tan60° = (H-h)/d (if hill is above tower... wait.

Actually: Tower top is observer. Hill foot is below (depression=30°).
tan30° = h/d → d = h√3.
Elevation to hill top from tower top = 60°.
tan60° = (H-h)/d? Only if hill top is above tower top.
(H-h) = d×tan60° = h√3×√3 = 3h.
H = h+3h = 4h.

But wait: if hilltop elevation=60° means hilltop is above observer (tower top):
Hill height - tower height = d×tan60° = h√3×√3 = 3h.
H = h+3h = 4h.

60°

tan θ = 20/(20/√3) = 20×√3/20 = √3.
θ = 60°.

Height = √3/2 km = 500√3 m

Let A closer, B farther. Height=h, A distance=d.
tan60°=h/d → d=h/√3.
tan30°=h/(d+1) → d+1=h√3.
h/√3+1=h√3 → 1=h√3-h/√3=h(√3-1/√3)=h(3-1)/√3=2h/√3.
h=√3/2 km = 866 m.