Heights & Distances (ఎత్తు మరియుదూరం )

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.

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 = 40/√3 = 40√3/3 m

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

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.

Hill = 150 m; Distance = 50√3 m

tan60°=h_hill/d → h_hill=d√3.
tan30°=50/d → d=50√3.
h_hill = 50√3×√3 = 150 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.

Height = 20 m

tan 45° = h/20 = 1.
h = 20 m.

45° angle → height = distance. Always!
ఇది exam లో quickest calculation!

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.

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).

L = h(tanθ₁/tanθ₂ + 1)

Horizontal distance d = h/tanθ₂.
Lighthouse top: above cliff observer by d×tanθ₁.
Height above observer = d tanθ₁ = h tanθ₁/tanθ₂.
Total lighthouse height = h + h tanθ₁/tanθ₂ = h(1+tanθ₁/tanθ₂).

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.

Distance = 8/√3 = 8√3/3 m

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

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 = 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.

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.

AT = 30√3/(√3-1) = 15√3(√3+1) = 45+15√3

Tower height h = AT (since 45°). AT = h.
From B: h/(AT-30) = tan60° = √3.
h/(h-30) = √3.
h = √3h-30√3.
h(1-√3) = -30√3.
h = 30√3/(√3-1) = 30√3(√3+1)/2 = 15√3(√3+1) = 45+15√3 m.

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 = 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!

Tower = 150 m

Let tower=H, pole=50m, distance=d.
From foot of pole to tower top: tan60°=H/d → H=d√3.
From foot of tower to pole top: tan30°=50/d → d=50√3.
H = 50√3×√3 = 150 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 = 30√3 m

tan 30° = 30/d → 1/√3 = 30/d → d = 30√3 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!

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!

Distance = 10√3 + 10/√3 = 40√3/3 m

Upstream boat: d₁ = 10/tan30° = 10√3.
Downstream boat: d₂ = 10/tan60° = 10/√3 = 10√3/3.
Total = 10√3+10√3/3 = 30√3/3+10√3/3 = 40√3/3 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.

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.

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!

AC = 30(√3+1)/(√3-1)... let me solve

Let C నుండి tower base D = d.
Tower top = B (30m), bottom = A (ground level? or B is base).
Actually: AB is a tower, A base, B top. Height AB=30m.
Elevation to B = 45°: tan45°=30/d → d=30.
Elevation to A from C = 30°: tan30°=something.

Actually: A is at height h₁, B at h₂=h₁+30.
Let's say A is some point on tower, B is top.
elevation to A = 30°, elevation to B = 45°, height diff = 30.
Let ground to A = h, ground to B = h+30.
tan45° = (h+30)/d → d = h+30.
tan30° = h/d → h/d = 1/√3 → h = d/√3 = (h+30)/√3.
h√3 = h+30 → h(√3-1) = 30 → h = 30/(√3-1) = 15(√3+1).
d = h+30 = 15(√3+1)+30 = 15√3+15+30 = 15√3+45 = 15(√3+3).

AC = d/cos(some angle)... AC=d (horizontal) = 15(√3+3).

Sun angle α from horizontal. Slope β. Shadow on slope = h sinα/sin(α+β)=h sinα/sin90°=h sinα

If α+β=90°: sin(α+β)=sin90°=1.
Shadow on slope = h sinα/sin(α+β) = h sinα/1 = h sinα.
But tanα = sinα/cosα. If β=90°-α:
Shadow = h sin α. When α=45°: shadow=h/√2 on the slope.

Using sine rule in the shadow triangle: shadow/sinα = h/sin(90°-β)=h/sinα...
shadow = h sinα/sin(α+β) = h sinα (when α+β=90°).

క్షితిజ రేఖ నుండి పైకి చూసే కోణం

Observer horizontal line నుండి పైన ఉన్న object వైపు చూసే కోణం = Angle of Elevation.
ఉదా: నేల పై నిలబడి tower top చూస్తే → angle of elevation.

గుర్తు: Elevation = పైకి చూడడం (ఆకాశం వైపు).
Depression = కిందికి చూడడం (నేల వైపు).
రెండూ horizontal line తో కొలుస్తాయి!

This is impossible — same height buildings have 0° to each other's top if same level.

Let tower height = h, building height = h, distance = d.
From top of tower to top of building: elevation means building top is above tower top → buildings not same height.

Re-read: From tower base, building top elevation=30°. From building base, tower top elevation=60°.
Tower height H, Building height h.
tan60° = H/d → H = d√3.
tan30° = h/d → h = d/√3.
H/h = 3. Buildings NOT same height.

Standard version: tan60°=H/d, tan30°=h/d. H:h = 3:1.

Height = 150/(√3-1) = 75(√3+1) m

Distance moved = 50×3 = 150 m.
At 45°: d=h. At 30°: d+150=h√3.
h+150=h√3 → h(√3-1)=150 → h=150/(√3-1)=75(√3+1) ≈ 204.9 m.

Lamp height = 7.8 m

Shadow tip is 10√3+6 m from lamp.
By similar triangles: h/(10√3+6) = 1.8/6.
h = 1.8×(10√3+6)/6 = 0.3×(10√3+6) = 3√3+1.8 ≈ 5.196+1.8 = 6.996 ≈ 7 m.

Exact: 3√3+1.8. Approx ≈ 7 m.

45°

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

Height = shadow → always 45°!

Height = x√3/2

tan30°=h/d → d=h√3.
tan60°=h/(d-x) → d-x=h/√3.
h√3-x=h/√3 → h√3-h/√3=x → h(3-1)/√3=x → 2h/√3=x → h=x√3/2.

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].

30°

tan θ = height/shadow = 1.5/(2√3) = 1.5/(2√3) = 3/(4√3) = 3√3/12 = √3/4.
Hmm not standard.

1.5 m = 3/2 m. tan θ = (3/2)/(2√3) = 3/(4√3) = √3/4.
Not standard.

Use boy=1.5m=√3m: tan θ=√3/(2√3)=1/2. Not standard.

Use shadow=1.5√3: tanθ=1.5/1.5√3=1/√3 → 30°.

Use boy=1.5m, shadow=1.5√3m: tanθ=1/√3 → 30°.
OR boy=150cm=1.5m, shadow=2.6m≈1.5√3: tanθ=1/√3→30°.

Find h first, then compute θ₃

h = 10√3/(√3-1) = 5√3(√3+1) = 15+5√3.
After 10m: d = h (from 45°). After 10m: d₁=h-10.
After 15m: d₂=h-15.
tan θ₃ = h/(h-15) = (15+5√3)/(15+5√3-15) = (15+5√3)/(5√3) = 3/√3+1 = √3+1.
θ₃ = arctan(√3+1) ≈ arctan(2.732) ≈ 69.9°.

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.

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.

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.

Tower A = 30√3 m (taller)

Tower A: h_A = 30×tan60° = 30√3 ≈ 51.96 m.
Tower B: h_B = 40×tan45° = 40 m.
Taller = Tower A = 30√3 ≈ 51.96 m > 40 m.

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!

Shadow rate = 2/3 m/s ✓

Let x = man distance from lamp, s = shadow length.
By similar triangles: 5/(x+s) = 1.6/s → 5s = 1.6x+1.6s → 3.4s = 1.6x → s = 1.6x/3.4 = 8x/17.
ds/dt = (8/17)×dx/dt = (8/17)×1 = 8/17 ≈ 0.47 m/s.

Hmm 8/17 ≠ 2/3. Let me use man=2m, lamp=6m:
6/(x+s)=2/s → 6s=2x+2s → 4s=2x → s=x/2. ds/dt=½ m/s.

For 2/3: lamp=5, man=2: 5/(x+s)=2/s → 5s=2x+2s → 3s=2x → s=2x/3. ds/dt=2/3 ✓!
Man height=2m.

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.

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!

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!

Flag = 9(√3-1/√3) = 9×(2/√3) = 6√3 m

Tower height = 9×tan30° = 9/√3 = 3√3.
Total height = 9×tan60° = 9√3.
Flag = 9√3-3√3 = 6√3 m.

Shadow = h cotθ. Change = h cotβ-h cotα = h(cotβ-cotα) ✓

At angle α: shadow = h/tanα = h cotα.
At angle β: shadow = h cotβ.
Change = b-a = h cotβ-h cotα = h(cotβ-cotα). ✓

(β < α means sun lower → longer shadow → cotβ > cotα → b > a ✓)

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!