Applications — heights and distances problems

2.4.7 · Maths › Trigonometry — Foundation

Heights and distances problems trigonometric ratios use karte hain unknown lengths ya angles find karne ke liye jab direct measurement possible nahi hoti. Ye woh reason hai kyun trigonometry exist karti hai — ancient surveyors ko pyramid ki heights measure karni thi, sailors ko stars se navigate karna tha, aur engineers ko rivers cross kiye bina bridge spans calculate karne the.

[!intuition] Core Idea

Tum ek building ko ground se dekh rahe ho. Tum measure kar sakte ho:

• Building se apni distance (horizontal)
• Wo angle jo tumhari aankhein horizontal ke saath banati hain (angle of elevation)

Tum building ki height directly measure karne ke liye upar nahi chadh sakte.

Trigonometry angles ko lengths mein convert karne deti hai. Agar tum ek right triangle ki ek side aur ek angle jaante ho, toh ratios (tan, sin, cos) baaki sides unlock kar dete hain.

YE KYU KAAM KARTA HAI: Angle of elevation triangle ki shape determine karti hai. Jab shape fix ho jaati hai, agar tum ek side ko scale karo (jo distance tumne measure ki), toh baaki saari sides proportionally scale hoti hain. Tan, sin, cos wohi scale factors hain.

[!definition] Key Terms

Angle of elevation: Horizontal line (apni eye level se) aur line of sight ke beech ka angle jo upar tumse upar kisi object ki taraf jaata hai.

Angle of depression: Horizontal line aur line of sight ke beech ka angle jo neeche tumse neeche kisi object ki taraf jaata hai (jaise cliff ke upar se neeche ek boat ko dekhna).

YE DONO OPPOSITE VIEWPOINTS SE EQUAL KYU HAIN: Agar tum ground par angle α se upar dekh rahe ho, toh upar wala banda tumhe neeche dekh ke angle α hi dekhega — ye alternate interior angles hain. Tumhari eye level par horizontal line aur upar ki horizontal line do parallel lines hain, aur line of sight woh transversal hai jo dono ko cut karti hai. Vertical height sirf do lines ke beech ka segment hai, transversal nahi.

Line of sight: Tumhari aankhon se object tak seedhi line. Ye tumhare right triangle ki hypotenuse ban jaati hai.

Horizontal distance: Tumse object ke base tak ground-level distance. Hamesha vertical height ke perpendicular hoti hai.

[!formula] Basic Setup Derive Karna

GIVEN: Tum point A par khade ho, tower ke top B ko dekh rahe ho. Tower ka base C hai. Tum measure karte ho:

• Horizontal distance AC = $d$
• Angle of elevation ∠BAC = $\theta$

FIND: Tower ki height BC = $h$

Step-by-step derivation:

Step 1: Right triangle identify karo.

• Tower vertical hai → ∠ACB = 90°
• Triangle ABC, C par right-angled hai

Step 2: Sahi trig ratio choose karo. Hamare paas adjacent side hai (AC = $d$) aur hum opposite side (BC = $h$) chahte hain. $\tan \theta =\frac{\text{opposite}}{\text{adjacent}}=\frac{h}{d}$

TAN KYU? Kyunki tan triangle ke dono legs ko relate karta hai. Sin aur cos mein hypotenuse (line of sight) aati hai, jo na hum jaante hain na zaroori hai.

Step 3: Unknown solve karo. $h=d\tan \theta$

Physical meaning: Agar angle double ho jaaye, toh kya height double ho jaayegi? NAHI — tan nonlinear hai. $\tan 60°=\sqrt{3}\approx 1.73$, lekin $\tan 30°=1/\sqrt{3}\approx 0.58$. Jab $\theta \to 90°$, $\tan \theta$ without bound badhta hai (iska vertical asymptote hota hai), toh calculated height infinity ki taraf shoot karti hai. Note karo ye growth exponential nahi hai ($e^{k\theta }$) — ye 90° par apne asymptote ki taraf tangent function ki specific unbounded growth hai.

[!example] Example 1: Basic Tower

Problem: Ek student tower se 50 m door khada hai. Top tak angle of elevation 30° hai. Tower ki height find karo.

Solution:

Given: d = 50 m, θ = 30°
Find: h

$h=d\tan \theta =50\times \tan 30°$

YE STEP KYU? Tan known adjacent (50 m) ko unknown opposite (h) se connect karta hai.

$h=50\times \frac{1}{\sqrt{3}}=\frac{50}{\sqrt{3}}=\frac{50\sqrt{3}}{3}$

RATIONALIZE KYU? Standard form ke liye. Numerically: $h\approx 28.87$ m.

Verification instinct: Kya ye sense karta hai? Angle chhota hai (30°), toh tower base distance (50 m) se chhotra hona chahiye. ✓ Check: 28.87 < 50.

[!example] Example 2: Two Observers (Distance Between)

Problem: 75 m cliff ke top se, ek boat ka angle of depression 30° hai. Boat cliff ke base se kitni door hai?

Solution:

Given: h = 75 m (cliff height), θ = 30° (depression angle)
Find: d (horizontal distance)

CRITICAL INSIGHT: Draw karo! Top se angle of depression, boat se angle of elevation ke equal hota hai (alternate angles).

Cliff-top se neeche dekhte hue: $\tan 30°=\frac{h}{d}=\frac{75}{d}$

YE SETUP KYU? Cliff height (75 m) angle of depression ke opposite hai, boat ki distance adjacent hai.

$d=\frac{75}{\tan 30°}=75\times \sqrt{3}=75\sqrt{3}$

Numerically: $d\approx 129.9$ m.

Verification: Angle 30° hai (chhota) → distance height se zyada honi chahiye. ✓ Check: 129.9 > 75.

[!example] Example 3: Change in Angle (Moving Observer)

Problem: Ek aadmi tower ke top ko 30° ke angle of elevation par observe karta hai. Woh 20 m paas aata hai; ab angle 60° hai. Tower ki height find karo.

Solution:

Let tower height = h
Initial distance from tower = x
After walking closer: distance = x - 20

Pehli position se: $\tan 30°=\frac{h}{x}⟹\frac{1}{\sqrt{3}}=\frac{h}{x}⟹x=h\sqrt{3}\text{..(1)}$

YE STEP KYU? Hum unknown distance $x$ ko $h$ ke terms mein express kar rahe hain, jo hum find karna chahte hain.

Doosri position se: $\tan 60°=\frac{h}{x-20}⟹\sqrt{3}=\frac{h}{x-20}⟹x-20=\frac{h}{\sqrt{3}}\text{..(2)}$

Equation (2) ko equation (1) se subtract karo: $x-(x-20)=h\sqrt{3}-\frac{h}{\sqrt{3}}$

SUBTRACT KYU? $x$ eliminate ho jaata hai, sirf $h$ bachta hai.

$20=h\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)=h\left(\frac{3-1}{\sqrt{3}}\right)=h\cdot \frac{2}{\sqrt{3}}$

$h=20\times \frac{\sqrt{3}}{2}=10\sqrt{3}\approx 17.32\text{ m}$

Physical intuition check: Woh 20 m paas aaya aur angle badh gaya (30° → 60°), jo sense karta hai — tower ki taraf move karne se woh zyada tall dikhta hai (steeper line of sight). Height (17.32 m) kitne door woh chala (20 m) ke comparable hai, jo reasonable hai.

[!example] Example 4: Two Angles from a Fixed Point

Problem: Ground par ek point se, ek building par lage flagpole ke bottom ka angle of elevation 30° hai, aur flagpole ke top ka 45°. Agar building 20 m tall hai, toh flagpole ki height find karo.

Solution:

Building height = 20 m
Let flagpole height = f
Let horizontal distance from observer to building = d

YE SETUP CONSISTENT KYU HAI (kharab-posed problem ke unlike): Yahan dono angles same fixed point se same vertical line par do alag heights tak measure kiye gaye hain. Iska hamesha ek clean solution hota hai.

Flagpole ke bottom (building ke top) tak angle, 30°: $\tan 30°=\frac{20}{d}⟹\frac{1}{\sqrt{3}}=\frac{20}{d}⟹d=20\sqrt{3}\text{ m}\text{..(1)}$

KYU? Building ka top (20 m) 30° angle ke opposite hai; $d$ adjacent hai.

Flagpole ke top tak angle, 45°: $\tan 45°=\frac{20+f}{d}⟹1=\frac{20+f}{d}⟹20+f=d\text{..(2)}$

(1) ko (2) mein substitute karo: $20+f=20\sqrt{3}$ $f=20\sqrt{3}-20=20(\sqrt{3}-1)\approx 20\times 0.732=14.64\text{ m}$

Verification: Angle 30° se 45° badha jab hum upar dekhe, toh total height (34.64 m) building (20 m) se zyada honi chahiye. ✓ Aur $34.64=20\sqrt{3}=d$, 45° condition match karta hai (45° par, height = distance). ✓

Steel-man note: Ek tempting lekin ill-posed version kehta hai "ek bird ka angle of depression 45° hai, phir 100 m uraane ke baad 30°." Agar tum assume karo ki bird door constant height par jaata hai, toh dono conditions force karti hain $d_2-d_1=100$ lekin geometry deta hai $d_2-d_1=100(\sqrt{3}-1)\approx 73\ne 100$ — koi consistent solution nahi. Answer trust karne se pehle hamesha check karo ki diya gaya displacement jo angles imply karte hain usse match karta hai.

[!mistake] Common Mistakes

Mistake 1: Adjacent aur opposite confuse karna

Galat soch: "Tower mere paas hai, toh ye adjacent hai. Main $\sin \theta =h/d$ use karunga."

YE SAHI KYU LAGTA HAI: Tower physically tumhare paas hai.

Fix: Adjacent/opposite angle ke relative hote hain, tumhare physical position ke nahi. Agar angle us base par hai jahan tum khade ho, toh tower (vertical) angle ke opposite hai. Ground distance angle ke adjacent hai.

Memory aid: Triangle draw karo. Angle label karo. Angle ke saamine wali side opposite hai.

Mistake 2: Formula values ki jagah degrees use karna

Galat: $\tan 30°=30$ (angle ko directly number ki tarah treat karna).

YE SAHI KYU LAGTA HAI: Simple arithmetic mein hum numbers directly use karte hain.

Fix: Trig ratios ratios hain, angles themselves nahi. $\tan 30°=1/\sqrt{3}\approx 0.577$, 30 nahi.

Standard values use karo:

• $\tan 30°=1/\sqrt{3}$
• $\tan 45°=1$
• $\tan 60°=\sqrt{3}$

Mistake 3: Ye bhool jaana ki angle of depression = angle of elevation (alternate angles)

Galat setup: Angle of depression ko top par triangle ke andar angle ki tarah draw karna.

YE SAHI KYU LAGTA HAI: Observer top par hai, toh angle wahan hona chahiye.

Fix: Angle of depression observer ki eye level par horizontal se measure hota hai. Jab tum ise triangle mein transfer karte ho, alternate interior angles use karo: dono horizontal lines parallel hain, line of sight transversal hai, toh top par depression angle base par elevation angle ke equal hai.

Visual check: Do parallel horizontal lines draw karo (ek cliff top par, ek ground par). Line of sight dono ko cut karne wali transversal hai. Alternate interior angles equal hote hain.

Mistake 4: Units ya reasonableness check na karna

Galat: Height = 500 m calculate karo 50 m door se 10° angle ke liye, aur accept kar lo.

YE SAHI KYU LAGTA HAI: Tum apni algebra par trust karte ho.

Fix: $\tan 10°\approx 0.176$. Height = $50\times 0.176\approx 8.8$ m, 500 m nahi. Hamesha estimate karo: chhote angles → height < base distance.

[!recall]- Ek 12-year-old ko explain karo

Socho tum jaanna chahte ho ki ek bade daraKht ki height kitni hai, lekin tum usse chadh nahi sakte. Trick ye hai:

Kuch door khade ho aur top ko dekho. Tumhari aankhein ground ke saath ek angle banati hain — yahi angle of elevation hai. Ye aise hai jaise apna sar upar tilt karna.

Ab tumne ek invisible triangle bana liya hai: ground ek side hai (ise tape se measure kar sakte ho), darakht doosri side hai (yehi tum chahte ho), aur treetop tak tumhari line of sight slanted side hai.

Ye magic hai: us angle ka tan sirf ek ratio hai. Ye tumhe batata hai: "Har meter ki door se, darakht tan(angle) meter tall hai."

Toh agar tum 10 meter door ho aur angle 45° hai, tan(45°) = 1, toh darakht 10 × 1 = 10 meter tall hai. Agar angle 60° hota (steeper), tan(60°) ≈ 1.7, toh darakht 10 × 1.7 = 17 meter hota!

Ye kaam kyu karta hai? Kyunki angles triangles ki shape lock kar dete hain. Jab tum shape jaante ho (angle se) aur ek side ka size (apni distance), tum baaki saari sides figure out kar sakte ho in special ratios se jo hum sine, cosine, aur tangent kehte hain.

[!mnemonic] Memory Aid

"STAND-SEE-CLIMB"

• STAND: Apni distance measure karo (angle ke adjacent)
• SEE: Angle of elevation measure karo (upar dekho)
• CLIMB: Height calculate karo (opposite side) $h=d\cdot \tan \theta$ use karke

Angle of depression ke liye: "Jo main yahan se NEECHE dekh raha hoon = Jo woh mujhe UPAR dekhte hain" (alternate angles, line of sight transversal hai).

Connections

• 2.4.01-Trigonometric-ratios — Sin, cos, tan ki definitions jo in saari calculations ko power deti hain
• 2.4.03-Complementary-angles — Kyun angle of depression, angle of elevation ke equal hota hai
• 2.5.02-Surveying-and-navigation — In techniques ke real-world applications
• 3.2.04-Similar-triangles — Kyun angle shape determine karta hai, in measurements ko enable karta hai
• 1.3.07-Pythagoras-theorem — Tab use hota hai jab hypotenuse chahiye hो (line of sight distance)

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Deep Dive

• D1 · Foundations — har symbol zero se
• D2 · Visual walkthrough — derivation pictures mein
• D3 · Worked examples — har scenario
• D4 · Exercises — graded, full solutions
• D5 · Question bank — concept traps