Class 10 Maths · Chapter 9
Short answer:
Chapter 9 me sirf ek exercise hai — Exercise 9.1, 16 questions — poora chapter word problems ka hai jisme Chapter 8 ke trig ratios (sin, cos, tan) ko real-life heights aur distances nikalne me use karte hain. Har question me ek right triangle chhupa hota hai: kabhi tower/building/pole ki height nikalni hai, kabhi observer se object ki horizontal distance, kabhi rope/string/slide/broken-tree ki length. Sabse important skill hai situation ko sahi diagram me convert karna — angle of elevation ya depression kis point se naapa gaya hai, aur kaunsi side given hai, kaunsi find karni hai.
Ye chapter Chapter 8 ke trigonometric ratios ka real-world use hai — bina naapi hui heights aur distances nikalna sirf angle aur ek known length se. Jaise ek surveyor bina tower pe chadhe uski height bata deta hai, ya ek pilot bina neeche jaaye zameen ke do points ke beech distance nikal leta hai — sab kuch right-triangle trigonometry se hota hai. Is chapter me naya trig nahi seekhna — bas Chapter 8 ke sin/cos/tan values (30°, 45°, 60°) ko sahi triangle pe apply karna seekhna hai.
Chapter 9 Summary — 5 Minute Revision
1. Angle of elevation aur angle of depression
Dono angles hamesha horizontal line se naape jaate hain — observer ki aankh (ya point) se horizontal line kheencho, phir line-of-sight tak ka angle napo.
- Angle of elevation: Jab observer horizontal se upar kisi object (jaise tower ka top) ko dekhta hai. Object observer se upar hai.
- Angle of depression: Jab observer horizontal se neeche kisi object (jaise zameen par khadi car) ko dekhta hai. Observer object se upar hai (jaise tower/building ke top pe khada hai).
Common confusion: angle of depression observer ki horizontal line se naapa jaata hai, ground se nahi. Ye rule alternate angles (parallel horizontal lines) se link hota hai — object se dekhne wala angle of elevation, observer ke angle of depression ke barabar hota hai (alternate interior angles, kyunki dono horizontal lines parallel hain).
2. Diagram banane ka standard tareeka
- Right triangle banao — right angle hamesha vertical object (tower/pole/tree) aur ground ke beech hota hai.
- Given angle ko sahi vertex pe likho (observer ke point pe, horizontal line se).
- Jo side given hai (height ya distance) usse label karo; jo find karni hai usse variable do (h ya x).
- Sahi trig ratio choose karo — opposite/adjacent decide karta hai tan chahiye, sin, ya cos.
3. Kaunsa ratio kab use karo
| Given + Find | Ratio |
|---|---|
| Base distance (adjacent) given, height (opposite) find karni | tan θ = opposite / adjacent |
| Hypotenuse (rope/string/ladder length) given, height find karni | sin θ = opposite / hypotenuse |
| Hypotenuse given, base distance find karni | cos θ = adjacent / hypotenuse |
↔ Table ko side me swipe karein
4. Standard angle values (Chapter 8 se recap)
| θ | sin θ | cos θ | tan θ |
|---|---|---|---|
| 30° | 1/2 | √3/2 | 1/√3 |
| 45° | 1/√2 | 1/√2 | 1 |
| 60° | √3/2 | 1/2 | √3 |
↔ Table ko side me swipe karein
Ye chapter sirf 30°, 45°, 60° use karta hai — koi naya angle nahi. Answers usually surd form me chhodte hain (jaise 10√3 m), decimal sirf tab jab NCERT khud deta ho.
5. Do-observer / two-triangle problems
Jab do angles diye ho do alag points se (ya ek point pehle-baad me move karta ho), to do equations banao (dono triangles se), phir unhe solve karo. Common trick: ek unknown (height ya distance) common rehta hai dono equations me — usse eliminate karke doosra unknown nikalo.
6. Complementary angle result (Q16 ka general case)
Agar do points se angles of elevation complementary hain (θ aur 90°−θ), aur distances a, b hain object ke foot se, to height h = √(ab) — ye ek proof-type result hai, formula ratta mat lagao, derivation samjho.
Poore Class 10 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q16)
Q1. Ek circus artist 20 m lambi rope pe climb karta hai, jo ek vertical pole se lagi hai aur ground se 30° ka angle banati hai. Pole ki height nikalo, agar artist rope ke top tak pahunchta hai (rope ko slack maan kar mat lo).
Pole ki height = 10 m.
Rope = hypotenuse (20 m), angle 30°, height = opposite side.
sin 30° = height / rope = h / 20
h = 20 × sin 30° = 20 × (1/2) = 10 m
Q2. Ek storm ke kaaran ek ped bich se toot gaya. Ped ke tute hue hisse ka top ground ko chuta hai aur ground ke saath 30° ka angle banata hai. Jis point pe top ground ko chuta hai, wo ped ke root se 8 m door hai. Ped ki original height nikalo.
Ped ki original height = 8√3 m ≈ 13.86 m.
Do parts: (1) standing part (vertical, height = h₁), (2) broken part (jhuka hua, ground ko 30° pe chhoo raha, base = 8 m).
Standing part: tan 30° = h₁ / 8 ⇒ h₁ = 8/√3 = 8√3/3 m
Broken part (hypotenuse): cos 30° = 8 / (broken part) ⇒ broken part = 8/cos30° = 8/(√3/2) = 16/√3 = 16√3/3 m
Total original height = h₁ + broken part = 8√3/3 + 16√3/3 = 24√3/3 = 8√3 m
Q3. Ek contractor bachon ke liye ek slide banata hai. 5 saal se chhote bachon ke liye slide ka top ground se 1.5 m upar hai aur slide ground se 30° ka angle banata hai. Bade bachon ke liye ek steeper slide chahiye jiska top 3 m upar ho aur ground se 60° ka angle bane. Dono slides ki length nikalo.
Choti slide ki length = 3 m; badi slide ki length = 2√3 m ≈ 3.46 m.
Height = opposite side, slide length = hypotenuse.
Choti slide (1.5 m, 30°): sin 30° = 1.5 / L₁ ⇒ L₁ = 1.5 / (1/2) = 3 m
Badi slide (3 m, 60°): sin 60° = 3 / L₂ ⇒ L₂ = 3 / (√3/2) = 6/√3 = 2√3 m
Q4. Ek tower ke foot se 30 m door ek point hai, jahan se tower ke top ka angle of elevation 30° hai. Tower ki height nikalo.
Tower ki height = 10√3 m ≈ 17.32 m.
Base = 30 m (adjacent), height = opposite.
tan 30° = h / 30 ⇒ h = 30 × tan 30° = 30 × (1/√3) = 30/√3 = 10√3 m
Q5. Ek kite (patang) ground se 60 m ki height pe ud rahi hai. Kite ki string, horizontal se 60° ka angle banati hai. String ki length nikalo, assume karo koi slack nahi hai.
String ki length = 40√3 m ≈ 69.28 m.
Height = opposite (60 m), string = hypotenuse.
sin 60° = 60 / L ⇒ L = 60 / (√3/2) = 120/√3 = 40√3 m
Q6. Ek 1.5 m lamba ladka ek 30 m ooncchi building se kuch distance pe khada hai. Uski aankhon se building ke top ka angle of elevation 30° hai; jab wo building ki taraf chalta hai to angle 60° ho jaata hai. Ladka kitni distance chala, building ki taraf?
Ladka 19√3 m ≈ 32.91 m chala.
Ladke ki eye-level height 1.5 m hai, isliye effective height (eye-level se top tak) = 30 − 1.5 = 28.5 m — dono triangles me same rahega.
Pehli position (30°): tan 30° = 28.5 / d₁ ⇒ d₁ = 28.5 × √3 = 28.5√3 m
Doosri position (60°): tan 60° = 28.5 / d₂ ⇒ d₂ = 28.5/√3 = 28.5√3/3 = 9.5√3 m
Distance chala = d₁ − d₂ = 28.5√3 − 9.5√3 = 19√3 m
Q7. Ek 20 m ooncchi building ke top par ek transmission tower fix hai. Ground ke ek point se building ke bottom ka angle of elevation 45° hai aur tower ke top ka angle of elevation 60° hai. Tower ki height nikalo.
Tower ki height = 20(√3 − 1) m ≈ 14.64 m.
Point se building ka base distance d: tan 45° = 20/d ⇒ d = 20 m (kyunki tan45°=1)
Building + tower (top) ka angle 60°: tan 60° = (20 + h)/d = (20+h)/20 = √3
20 + h = 20√3 ⇒ h = 20√3 − 20 = 20(√3 − 1) m
Q8. Ek 1.6 m lambi statue ek pedestal (base) ke top par khadi hai. Ground ke ek point se statue ke top ka angle of elevation 60° hai aur pedestal ke top ka angle of elevation 45° hai. Pedestal ki height nikalo.
Pedestal ki height = 0.8(√3 + 1) m ≈ 2.18 m.
Pedestal height x, distance d.
tan 45° = x/d ⇒ d = x (kyunki tan45°=1)
tan 60° = (x + 1.6)/d = (x+1.6)/x = √3
x + 1.6 = √3·x ⇒ x(√3 − 1) = 1.6 ⇒ x = 1.6/(√3−1)
Rationalise: x = 1.6(√3+1) / [(√3−1)(√3+1)] = 1.6(√3+1)/2 = 0.8(√3+1) m ≈ 0.8×2.732 ≈ 2.18 m
Q9. Ek straight highway ek tower ke foot tak jaati hai. Tower ke top se khada aadmi ek car dekhta hai jo uniform speed se tower ki taraf aa rahi hai — angle of depression 30° hai. 6 second baad angle of depression 60° ho jaata hai. Car ko is point se tower ke foot tak pahunchne me kitna time lagega?
Car ko is point se tower tak pahunchne me 3 second lagenge.
Tower height h maan lo (final answer h pe depend nahi karta).
Pehla point: tan 30° = h/d₁ ⇒ d₁ = h/tan30° = h√3
Doosra point (6 sec baad): tan 60° = h/d₂ ⇒ d₂ = h/tan60° = h/√3
6 second me tay ki gayi distance = d₁ − d₂ = h√3 − h/√3 = (3h−h)/√3 = 2h/√3
Speed = (2h/√3) / 6 = h/(3√3)
Bacha hua distance (d₂) tower tak = h/√3
Time = d₂ / speed = (h/√3) ÷ (h/(3√3)) = (h/√3) × (3√3/h) = 3 second
Q10. Ek 80 m chaudi road ke dono taraf equal height ke do poles khade hain. Road ke ek point se — jo dono poles ke beech hai — ek pole ke top ka angle of elevation 60° hai aur doosre ka 30° hai. Poles ki height aur point ki dono poles se distance nikalo.
Poles ki height = 20√3 m ≈ 34.64 m. Point pehle pole se 20 m aur doosre pole se 60 m door hai.
Point ki pehle pole se distance x, doosre se (80 − x). Dono poles ki height h same hai.
tan 60° = h/x ⇒ h = x√3
tan 30° = h/(80−x) ⇒ h = (80−x)/√3
Equate: x√3 = (80−x)/√3 ⇒ 3x = 80 − x ⇒ 4x = 80 ⇒ x = 20 m, (80−x) = 60 m
h = 20√3 m
Q11. Ek canal ke ek bank par ek TV tower vertically khada hai. Canal ke doosre bank ke ek point se — jo tower ke exactly opposite hai — tower ke top ka angle of elevation 60° hai. Isi line pe, is point se 20 m aur peeche ek doosre point se angle of elevation 30° hai. Tower ki height aur canal ki chaudai nikalo.
Tower ki height = 10√3 m ≈ 17.32 m; canal ki chaudai = 10 m.
Canal chaudai (opposite point se tower ke foot tak) = d, tower height = h.
tan 60° = h/d ⇒ h = d√3
Peeche wale point (d+20 door) se: tan 30° = h/(d+20) ⇒ h = (d+20)/√3
Equate: d√3 = (d+20)/√3 ⇒ 3d = d + 20 ⇒ 2d = 20 ⇒ d = 10 m
h = 10√3 m
Q12. Ek 7 m ooncchi building ke top se ek cable tower ke top ka angle of elevation 60° hai aur cable tower ke foot ka angle of depression 45° hai. Tower ki height nikalo.
Cable tower ki height = 7(1 + √3) m ≈ 19.12 m.
Building-tower horizontal distance d, tower height H.
Angle of depression to foot = 45°: tan 45° = 7/d = 1 ⇒ d = 7 m
Angle of elevation to top (building ke top se, extra height H−7): tan 60° = (H−7)/d = (H−7)/7 = √3
H − 7 = 7√3 ⇒ H = 7 + 7√3 = 7(1+√3) m ≈ 7 × 2.732 ≈ 19.12 m
Q13. Ek 75 m ooncchi lighthouse ke top se, do ships (jo lighthouse ke ek hi taraf hain) ke angles of depression 30° aur 45° hain. Do ships ke beech ki distance nikalo.
Do ships ke beech ki distance = 75(√3 − 1) m ≈ 54.9 m.
Lighthouse height 75 m. Angle of depression = angle of elevation (alternate angles), same as horizontal distance calc.
Farther ship (30°): tan 30° = 75/d₁ ⇒ d₁ = 75/tan30° = 75√3 m
Nearer ship (45°): tan 45° = 75/d₂ ⇒ d₂ = 75 m
Distance between ships = d₁ − d₂ = 75√3 − 75 = 75(√3 − 1) m ≈ 75 × 0.732 ≈ 54.9 m
Q14. Ek 1.2 m lambi ladki, ground se 88.2 m height pe horizontal line me chal rahi ek balloon dekhti hai. Kisi instant pe balloon ka angle of elevation uski aankhon se 60° hai. Kuch der baad angle 30° ho jaata hai. Us interval me balloon ne kitni distance tay ki?
Balloon ne 58√3 m ≈ 100.46 m distance tay ki.
Ladki ki eye-level height 1.2 m, isliye effective height = 88.2 − 1.2 = 87 m (dono positions me same rahega, kyunki balloon horizontal move karta hai).
Pehla instant (60°): tan 60° = 87/d₁ ⇒ d₁ = 87/√3 = 29√3 m
Doosra instant (30°): tan 30° = 87/d₂ ⇒ d₂ = 87/(1/√3) = 87√3 m
Distance tay ki = d₂ − d₁ = 87√3 − 29√3 = 58√3 m
Q15. Ek tower ke foot se ek building ke top ka angle of elevation 30° hai, aur building ke foot se tower ke top ka angle of elevation 60° hai. Agar tower ki height 50 m hai, to building ki height nikalo.
Building ki height = 50/3 m ≈ 16.67 m.
Building height H, tower height 50 m, horizontal distance d (dono ek hi ground level pe, ek doosre ke foot se dekh rahe).
Tower ke foot se building ke top ka angle 30°: tan 30° = H/d ⇒ d = H/tan30° = H√3
Building ke foot se tower ke top ka angle 60°: tan 60° = 50/d = √3 ⇒ d = 50/√3
Equate: H√3 = 50/√3 ⇒ H = 50/3 m ≈ 16.67 m
Q16. Ek tower ke foot se ek hi seedhi line me, do points same side pe hain jinki tower ke foot se distances a aur b hain (a < b). Agar tower ke top ke angles of elevation in do points se complementary hain, to prove karo ki tower ki height h = √(ab) hai.
Proved: h = √(ab).
Maan lo pehle point (distance a) se angle of elevation θ hai, to doosre point (distance b, jo farther hai) se angle (90° − θ) hoga, kyunki dono complementary hain.
Pehla: tan θ = h/a
Doosra: tan(90°−θ) = cot θ = h/b
Dono equations multiply karo: tan θ × cot θ = (h/a) × (h/b)
Left side = 1 (kyunki tanθ × cotθ = 1)
1 = h²/(ab) ⇒ h² = ab ⇒ h = √(ab)
Important Equations — Ek Nazar Me
| Term | Definition |
|---|---|
| Angle of elevation | Horizontal line se upar dekhne par banne wala angle (object observer se upar hai) |
| Angle of depression | Horizontal line se neeche dekhne par banne wala angle (observer object se upar hai) |
| Alternate angle rule | Object se angle of elevation = observer ke angle of depression (parallel horizontal lines ke alternate interior angles) |
↔ Table ko side me swipe karein
| θ | sin θ | cos θ | tan θ |
|---|---|---|---|
| 30° | 1/2 | √3/2 | 1/√3 |
| 45° | 1/√2 | 1/√2 | 1 |
| 60° | √3/2 | 1/2 | √3 |
↔ Table ko side me swipe karein
| Given side pair | Ratio use karo |
|---|---|
| Adjacent (base) known, opposite (height) find karni | tan θ = opposite/adjacent |
| Hypotenuse (rope/string/slide) known, opposite find karni | sin θ = opposite/hypotenuse |
| Hypotenuse known, adjacent find karni | cos θ = adjacent/hypotenuse |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Adjacent/opposite mix-up. Bahut students distance (base) ko opposite maan lete hain aur height ko adjacent — hamesha pehle diagram banao, right angle identify karo, phir decide karo kaunsi side opposite hai (angle ke saamne) aur kaunsi adjacent (angle se lagi hui, hypotenuse ko chhod kar).
- Angle of depression ko ground se naapna. Angle of depression observer ki horizontal line se naapa jaata hai, ground ya vertical tower se nahi. Diagram me observer ke point se ek horizontal line zaroor kheencho.
- Sahi trig ratio na choose karna. Jab hypotenuse (rope, string, ladder, slide) diya ho, tan ka use karna galat hai — sin ya cos use karo. tan sirf tab jab dono legs (opposite-adjacent) involve ho, hypotenuse nahi.
- Observer/object ki apni height ignore karna. Jab problem me observer ki height di ho (jaise '1.5 m lamba ladka' ya '1.2 m lambi ladki'), to effective height (tower/building ki height minus observer ki eye-level height) use karo, poori height nahi.
- Two-triangle problems me galat side subtract karna. Jab do angles do positions se diye ho, dono equations se distances nikal kar sahi order me subtract karo (bada distance − chota distance) — ulta karne se negative ya galat answer aata hai.
- Surd rationalise karna bhool jaana (ya inconsistently karna). 1/√3 ko √3/3 likhna sahi hai, lekin same answer ke andar kabhi rationalised kabhi nahi — ek hi convention poore solution me follow karo. NCERT usually exact surd form deta hai jab tak decimal specifically na maanga ho.
Board-Style Important Questions
- 1 mark: Angle of elevation aur angle of depression ki definition likho, ek diagram ke saath.
- 2 marks: Ek 15 m lambi ladder ek wall se is tarah lagi hai ki wo wall se 60° ka angle banati hai. Ladder ka foot wall se kitni door hai?
- 3 marks: Ek tower ke foot se 40 m door ek point se tower ke top ka angle of elevation 30° hai. Tower ki height nikalo.
- 3 marks: Do buildings same road ke opposite sides pe hain. Ek building (height 60 m) ke top se doosri building ke top aur foot ke angles of depression respectively 30° aur 60° hain. Doosri building ki height nikalo.
- 5 marks: Ek balloon horizontal line me udta hai. Ek observer, jo eye-level height h₀ pe hai, balloon ko pehle ek angle pe aur kuch der baad doosre angle pe dekhta hai. Given angles aur observer ki height se balloon ke travel ki distance nikalne ka poora method likho aur solve karo.
Quick Quiz — Score Check Karein
Q1. Angle of elevation kaise define hota hai?
Q2. Angle of depression kis line se naapa jaata hai?
Q3. Q1 (circus artist): 20 m lambi rope ek pole se lagi hai, ground se 30° angle banati hai. Pole ki height kya hai?
Q4. Kaunsi condition me 'sin θ = opposite/hypotenuse' use hoga (chapter ke table ke mutabik)?
Q5. Q4: Tower ke foot se 30 m door se tower ke top ka angle of elevation 30° hai. Tower ki height kya hogi?
Q6. Q5: Kite 60 m height pe hai, string horizontal se 60° angle banati hai. String ki length (no slack) kya hai?
Q7. Chapter 9 me kitne exercises hain?
Q8. Q16 ka general result: agar do points se angles of elevation complementary hain aur unki foot se distances a aur b hain, to tower ki height h kya hogi?
Q9. Is chapter me kaunse standard angles use hote hain?
Q10. Q9 (highway/car problem) me kaunsa concept use hota hai jab do angles of depression 6 second ke gap pe diye ho?
Aksar Poochhe Jaane Wale Sawaal
Angle of elevation aur angle of depression me kya farak hai?
Dono hamesha observer ki horizontal line se naape jaate hain. Elevation tab jab object upar hai (upar dekhna padta hai), depression tab jab object neeche hai (neeche dekhna padta hai, observer khud upar khada hai).
Is chapter me Chapter 8 ke formulas dobara seekhne padenge kya?
Naye formulas nahi hain — sirf Chapter 8 ke sin, cos, tan values (30°, 45°, 60° ke liye) ko real-life right-triangle problems pe apply karna hai. Agar wo values yaad nahi, pehle unhe revise kar lo.
Answer surd form me du ya decimal me?
NCERT usually exact surd form expect karta hai (jaise 10√3 m), jab tak question decimal specifically na maange. Dono forms sahi hoti hain lekin surd form zyada precise aur board-marking-safe hai.
Diagram banana zaroori hai kya, marks milte hain uske?
Diagram directly marks nahi deta lekin sahi diagram ke bina angle ka sahi placement, sahi ratio choose karna almost impossible hai — practically har question diagram ke bina galat ho jaata hai.
Kya iss chapter me koi naya trig identity seekhni hai?
Nahi. Sirf teen basic ratios (sin, cos, tan) aur teen standard angles (30°, 45°, 60°) ka application hai — Chapter 8 ka seedha extension.
Complementary angle wala proof question (Q16 type) common hai kya?
Ye ek conceptual/proof-type question hai jo is chapter ka general pattern samjhata hai — do observation points se complementary angles diye hon to height = √(ab). Isi tarah ke variations ban sakte hain, isliye method (do equations banake multiply karna) samajhna zaroori hai, sirf result ratta lagana kaafi nahi.
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