Heights & Distances (ఎత్తు మరియుదూరం )

Height = a√2 × tan60° = a√6

Diagonal of square = a√2.
tan60° = h/(a√2) → h = a√2×√3 = a√6 m.

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.

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).

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!

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!

50 m

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

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.

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 ✓)

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 = 50 m

Angle of depression = angle of elevation from boat = 45°.
tan 45° = cliff height/distance = 50/d.
1 = 50/d → d = 50 m.

Depression = Elevation (alternate angles). 45° → height = distance!

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°.

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.

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.

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.

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.

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.

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.

d_A = h/tanα, d_B = h/tanβ. Same distance → α=β. AB = d√2 = h√2/tanα

If same distance: d_A = d_B = d.
A is north, B is east → perpendicular directions.
AB = √(d²+d²) = d√2.
d = h/tanα (elevation from A = α).
AB = h√2/tanα = h√2 cotα.

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.

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.

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.

Distance = 100√3 m

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

Tower = 266.67 m = 800/3 m

Let d = horizontal distance, tower height = H.
Depression to tower base (depression = angle below horizontal from hill top):
tan60° = 200/d → d = 200/√3.
Elevation to tower top (above hill top? Below?):
If tower top is above hill top: tan30° = (H-200)/d → H-200 = d/√3 = 200/3.
H = 200+200/3 = 800/3 ≈ 266.7 m.

If tower top is below hill top: tower top depression, not elevation. Use elevation:
H > 200: H-200 = (200/√3)(1/√3) = 200/3. H = 200+200/3 = 800/3 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.

60°

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

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

60°

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

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.

Height = 3√3/4 km = 750√3 m

From A (60°): d_A = h/tan60° = h/√3.
From B (30°): d_B = h/tan30° = h√3.
d_A+d_B = 3 → h/√3+h√3 = 3 → h(1/√3+√3) = 3 → h(4/√3) = 3 → h = 3√3/4 km = 0.75√3 km ≈ 1.299 km = 1299 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!

Complex 3D navigation problem — set up coordinates

S₁=(0,0), L=(x,y) unknown. After 5km north: S₂=(0,5).
Bearing N30°E from S₁: L is at direction 30° east of north → x>0, y>0.
tan(bearing)=x/y. For N30°E: x/y=tan30°=1/√3 → y=x√3.
Elevation 20°: tanθ=h/√(x²+y²).
From S₂: bearing N60°E: x/(y-5)=tan60°=√3 → x=(y-5)√3.
Also y=x√3 → y=(y-5)√3×√3=3(y-5) → y=3y-15 → 2y=15 → y=7.5.
x=7.5/√3=2.5√3.
Distance from S₁=√(x²+y²)=√(18.75+56.25)=√75=5√3.
h=5√3×tan20°≈5√3×0.364≈3.15 km.

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°).

Height = 20 m

tan 45° = h/20 = 1.
h = 20 m.

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

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.

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.

h₁ = (w/2)tanα, h₂ = (w/2)tanβ

From midpoint (w/2 from each bank):
h₁ = (w/2)×tanα.
h₂ = (w/2)×tanβ.

Simple and elegant! Midpoint equidistant from both towers.

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.

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!

h = d tanα tanβ / (tanα - tanβ)

From B: h = (d_B)×tanβ. From A: h = (d_B+d)×tanα (A is farther).
(d_B+d)tanα = d_B tanβ.
d_B(tanβ-tanα) = d×tanα.
d_B = d tanα/(tanβ-tanα).
h = d_B×tanβ = d tanα tanβ/(tanβ-tanα).

Or: h = d/(cotα-cotβ) = d tanα tanβ/(tanβ-tanα).

Flagpole h = 20/√3 = 20√3/3 m

tan45°=20/d → d=20.
tan30°=h/d → h=d/√3=20/√3=20√3/3 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!

45°

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

Height = shadow → always 45°!

Height = 100/√3 = 100√3/3 m

tan 30° = h/100.
1/√3 = h/100.
h = 100/√3 = 100√3/3 ≈ 57.74 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.

Width = 20/√3 = 20√3/3 m

tan 60° = 20/width.
√3 = 20/w.
w = 20/√3 = 20√3/3 ≈ 11.55 m.

30°

tan θ = height/shadow = h/(h√3) = 1/√3.
tan θ = 1/√3 → θ = 30°.

shadow = h√3 → tan θ = 1/√3 → θ=30°.
గుర్తు: Shadow > height → small angle (30°); Shadow < height → large angle (60°+)!

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.

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 ✓

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].

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.