Some Applications of Trigonometry: Class 10 Maths Chapter 9 Hinglish mein

Ek length aur ek angle se un heights aur distances ko nikaalo jahan tum pahunch nahi sakte

Mathematics · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

50 min
24 practice questions
13 MCQs

Is chapter mein aap kya seekhenge

Kisi real situation mein line of sight, eye se guzarne wali horizontal, aur angles of elevation aur depression pehchaano
Samjhao ki upar se dekhe gaye point ka angle of depression, us point se dekhe gaye observer ke angle of elevation ke barabar kyun hota hai
Heights-and-distances problem ka right-triangle model banao aur woh trigonometric ratio chuno jo known aur unknown sides ko jodta hai
Ek right triangle se height, distance ya slanting length nikaalo, aur zaroorat ho to observer ki eye height jodo
Aise problems solve karo jinmein do right triangles ek side share karte hain, jaise ek point se dekhe gaye do angles ya do points se dekha gaya ek object
Answers surd form mein aur √3 ≈ 1.732 lekar decimals mein do, aur check karo ki woh sense banate hain

NCERT Mathematics Class 10, Chapter 9, pp. 133–143 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

9.1 Heights and Distances

Jis mandir mein tum jaate ho, uska gopuram kitna ooncha hai? Measuring tape lekar tum us par chadh nahi sakte, aur koi tumhe try bhi nahi karne dega. Surveyors, engineers aur sailors ke saamne yeh problem hamesha aati hai. Woh ground par ek length aur ek angle measure karte hain, aur baaki kaam trigonometry kar deti hai.

Chapter 8 mein tumne right triangle ke acute angle ke trigonometric ratios dekhe the. Ab tum unhe kaam mein laoge. Is chapter ka har problem ek hi plan follow karta hai: right triangle banao, jo pata hai use mark karo, aur woh ratio chuno jo use tumhari chahi gayi cheez se jodta hai.

Table 9.1Ratios jo tum baar-baar use karoge

	$30^{\circ}$	$45^{\circ}$	$60^{\circ}$
$\sin θ$	$\frac{1}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{\sqrt{3}}{2}$
$\cos θ$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{2}$
$\tan θ$	$\frac{1}{\sqrt{3}}$	$1$	$\sqrt{3}$
$\tan θ$ (decimal mein)	$0.577$	$1$	$1.732$

Board papers mein aksar likha hota hai take $\sqrt{3} = 1.732.$ Is poore chapter mein hum bhi yahi karenge. Surd form, jaise $13 \sqrt{3},$ ko aakhri line tak rakho aur tabhi use decimal mein badlo: jaldi round karne se answers khisak jaate hain.

Line of sight aur angle of elevation

Ek oonche tower se kuch door khade ho jao aur uske top ko dekho. Socho ki tumhari eye se us top tak ek dhaaga kheench kar tana hua hai. Yeh dhaaga tumhari line of sight (दृष्टि रेखा) hai.

Ab tumhari eye se ek doosra dhaaga socho, bilkul level, tower ki taraf point karta hua. Yeh tumhari eye se guzarne wali horizontal hai. Top dekhne ke liye tum sir upar uthaate ho, isliye line of sight horizontal se upar chali jaati hai. Dono dhaagon ke beech ka angle top ka angle of elevation (उन्नयन कोण) hai.

4/4Inke beech ka angle $θ$ angle of elevation hai. Top se aane wali vertical ke saath yeh ek right triangle banata hai.

Fig. 9.1Angle of elevation eye se guzarne wali horizontal se upar ki taraf measure hota hai.NCERT Fig. 9.1 · p. 133

Left mein seedhi-numa tiers wala ek mandir ka gopuram khada hai aur right mein level ground par ek person uski taraf munh karke khada hai. Step 1 sirf scene dikhata hai. Step 2 person ki eye se tower ki taraf ek dashed level line jodta hai, jis par horizontal likha hai. Step 3 eye se gopuram ke top tak ek solid line jodta hai, jis par line of sight likha hai. Step 4 eye par horizontal aur line of sight ke beech angle theta mark karta hai, us par angle of elevation likhta hai, aur top se eye level tak neeche aati dashed vertical line ke saath bane right triangle ko shade karta hai.

Definition

Line of sight

Woh seedhi line jo tumhari eye ko us point se jodti hai jise tum dekh rahe ho.

Definition

Angle of elevation

Jab dekha gaya point tumhari eye se ooncha ho, tab eye se guzarne wali horizontal se line of sight tak upar ki taraf measure kiya gaya angle.

ExampleJab tum patang, chhat ya Chaand ko upar dekhte ho, tab tum angle of elevation measure karte ho.

Do details mein log aksar galti karte hain. Angle horizontal se measure hota hai, tower ke paas wale ground se nahi aur vertical se bhi nahi. Aur yeh observer ki eye par measure hota hai, isliye ek hi jagah khade lambe aur chhote person ko thode alag angles dikhte hain.

Angle of depression

Elevation ki ek mirror image bhi hai. Jab dekha gaya point tumhari eye se neeche ho, to tum sir neeche jhukaate ho aur line of sight horizontal se neeche chali jaati hai. Jitna angle woh neeche girti hai, woh angle of depression (अवनमन कोण) hai. Socho ek lighthouse keeper gallery par khadi hai aur bahut neeche ek fishing boat ko dekh rahi hai: boat ka angle of depression uski eye par measure hota hai.

4/4Boat se guzarne wali horizontal uski horizontal ke parallel hai. Boat se keeper ka angle of elevation bhi wahi $θ$ hai: yeh alternate angles hain.

Fig. 9.2Angle of depression eye se guzarne wali horizontal se neeche ki taraf measure hota hai.NCERT Fig. 9.3 · p. 134

Right mein chattaanon par ek dhaariyon wala lighthouse khada hai jiski gallery par keeper hai; left mein samundar par ek chhoti sailing boat tair rahi hai. Step 1 scene dikhata hai. Step 2 keeper ki eye se samundar ke upar ek dashed horizontal line aur boat tak neeche ek solid line of sight jodta hai. Step 3 keeper ki eye par horizontal aur line of sight ke beech angle theta mark karta hai, jis par angle of depression likha hai. Step 4 boat se guzarti ek doosri dashed horizontal jodta hai, dono horizontals par chhote arrowheads lagaata hai taaki dikhe ki woh parallel hain, aur boat par barabar angle theta mark karta hai, jis par angle of elevation, boat se likha hai.

Definition

Angle of depression

Jab dekha gaya point tumhari eye se neeche ho, tab eye se guzarne wali horizontal se line of sight tak neeche ki taraf measure kiya gaya angle.

ExampleJab tum balcony se neeche street par khadi scooter ko dekhte ho, tab tum angle of depression measure karte ho.

Elevation ya depression?

4 mein se 1

Riya apni first-floor balcony par khadi hai aur neeche street par khadi scooter ko dekhti hai. Uski line of sight horizontal ke saath jo angle banati hai, woh hai

angle of depressionangle of elevationwall se measure kiya gaya angle

Answer dekho

Answer (A) angle of depression

Scooter uske eye level se neeche hai, isliye woh neeche dekhti hai: angle of depression. Elevation ka matlab upar dekhna hota, aur dono mein se koi bhi angle wall se measure nahi hota.

Heights problem ka plan banana

Tumhare school ground mein ek neem ka ped hai jo tape se naapne ke liye bahut ooncha hai. Aisi jagah khade ho jao jahan se uska top dikhe. Tumhari eye, ped ka top aur tane par tumhari eye ke level wala point ek right triangle banate hain. Ise solve karne ke liye tumhe do numbers chahiye: top ka angle of elevation aur tane se tumhari distance. Ek aur number kaam poora karta hai: ground se tumhari eye ki height, kyunki triangle wahin se shuru hota hai, ground se nahi.

3/3Ped $x$ se uski eye height jitna zyada ooncha hai, yaani $e,$ isliye $H = e + x = e + d \tan θ .$

Fig. 9.3Teen measurements se height milti hai: distance, angle aur eye height.NCERT p. 135

Left mein ek bada ped khada hai aur right mein ek student apni eye par ek chhota clinometer lagaye hue hai. Step 1 scene dikhata hai. Step 2 model banata hai: uski eye se ped ke tane tak ek dashed horizontal, ped ke top tak line of sight, top se neeche aati dashed vertical jis par right-angle mark hai, uski eye par angle theta, ground distance d, aur uski eye ke upar top ki height x, jo left mein mark hai. Step 3 uski eye height e jodta hai, jo ground se eye level tak mark hai, aur result H equals e plus d tan theta.

Fig. 9.3 ke triangle ka right angle tane par, tumhari eye ke level par hai. Tane ke saath wali side $x,$ angle ke opposite hai, jo hai $θ .$ Iski level side $d,$ bhi usi angle ko chhooti hai, yaani $θ .$ Tangent wahi ratio hai jo bilkul inhi do sides se banta hai, isliye $x = d \tan θ .$ Ismein apni eye height jodo aur tumhe ped ki height mil jaayegi.

Angle of elevation se height

H = e + d \tan θ

(9.1)

jahan

$H$: object ki height
$e$: ground se observer ki eye ki height (agar angle ground level par measure hua hai to 0)
$d$: observer se object ke foot tak horizontal distance
$θ$: top ka angle of elevation

Har problem height nahi poochhta. Kabhi slanting side pata hoti hai (ladder, rassi, patang ki dor), ya wahi cheez chahiye hoti hai. Woh ratio chuno jismein tumhari jaani hui side aur chahi gayi side ho, aur kuch nahi.

Table 9.2Kaunsa ratio use karein

Tumhe pata hai	Tumhe chahiye	Use karo
horizontal distance (adjacent)	height (opposite)	$\tan θ = \frac{opp}{adj}$
height (opposite)	horizontal distance (adjacent)	$\tan θ$
slanting side (hypotenuse)	height (opposite)	$\sin θ = \frac{opp}{hyp}$
height (opposite)	slanting side (hypotenuse)	$\sin θ$
slanting side (hypotenuse)	horizontal distance (adjacent)	$\cos θ = \frac{adj}{hyp}$

Chalo explore karein

Clinometer banao aur kisi oonchi cheez ko measure karo

Tumhe chahiye

ek protractor jiske centre mein chhota sa chhed ho
ek drinking straw aur sticky tape
20 cm dhaaga jiske ek end par chhota sa nut bandha ho
ek measuring tape
ek partner

Pehle guess karo

Tum school ke flagpole ka top straw se dekhte ho, phir das kadam peeche jaakar use dobara dekhte ho. Doosri baar jo angle padhoge, woh hoga

badachhotasame

Straw ko protractor ke seedhe edge ke saath tape kar do. Dhaage ko centre wale chhed se nikaalo taaki nut curved scale ke neeche aazaadi se latke.
Straw se flagpole ya ped ke top ko dekho. Hilo mat, jab tak tumhara partner padhe ki dhaaga scale ko kahan kaat raha hai.
Dhaaga hamesha seedha neeche latakta hai. Isliye angle of elevation reading aur $90^{\circ}$ ka difference hai: $60^{\circ}$ ya $120^{\circ}$ ki reading, dono ka matlab hai $30^{\circ} .$
Object ke foot se apni distance aur ground se apni eye ki height measure karo.
(9.1) se height nikaalo. Phir ek doosri jagah se dohrao.

Table 9.3Meri measurements

Jagah	$d$ (m)	$θ$ (°)	$H$ (m)
Pehli jagah
Doosri jagah

Tumhari entries apne aap save hoti hain.

InteractiveElevation explorer

Scale ke hisaab se bana ek interactive scene. Look up view mein left mein gopuram, lighthouse ya coconut palm jaisa ek ooncha object khada hai aur right mein ground par ek observer khada hai. Distance d badalne ke liye tum observer ko drag karte ho, ya slider use karte ho. Line of sight, eye se guzarne wali horizontal aur angle of elevation bane hote hain, aur ek panel d, angle, angle ka tan, eye ke upar ki height d tan theta aur total height e plus d tan theta dikhata hai. Buttons observer ko wahan le jaate hain jahan angle bilkul 30, 45 ya 60 degrees ho, ek switch 1.5 m ki eye height jodta hai, aur ek Play button observer ko object ki taraf chalata hai. Look down view mein tum ek boat drag karte ho aur dekhte ho ki top se angle of depression, boat se angle of elevation ke barabar hai. Challenge mode height chhupa deta hai aur tumse distance aur angle se use calculate karne ko kehta hai.

Fig. 9.4Observer ko move karo aur dekho ki angle of elevation aur

d \tan θ

kaise badalte hain.

Ratio chuno

3 mein se 1

Makar Sankranti par Ishaan ki patang ki dor 100 m lambi hai aur uske haath par horizontal ke saath

30^{\circ}

banati hai. Dor ko seedha maan kar batao, patang uske haath se kitni oonchi hai?

Answer dekho

Answer 50 m

Dor hypotenuse hai, isliye

\sin 30^{\circ} = \frac{h}{100}

aur

h = 100 \times \frac{1}{2} = 50 m .

\tan

use karne ke liye horizontal distance chahiye hoti, jo tumhe pata nahi hai.

Ek right triangle wale problems

Chalo plan use karte hain. Har example mein pehle figure ka sketch banao, phir right angle mark karo, aur uske baad hi ratio chuno.

1/2Point $P,$ jo 39 m door hai, wahan se line of sight $30^{\circ}$ par palm ke top tak uthti hai.

Fig. 9.5Example 9.1:

P

se palm ke top tak line of sight.

Level ground ke left mein ek lamba coconut palm khada hai jiske top par patte hain. Point P right mein ground par hai, palm ke foot se 39 m door, jo dimension arrow se mark hai. Step 1 P se palm ke top tak line of sight dikhata hai jo ground ke saath 30 degrees banati hai. Step 2 right triangle ko shade karta hai, foot par right-angle mark lagaata hai, unknown height h mark karta hai, aur result h equals 13 root 3, lagbhag 22.52 m dikhata hai.

Example 9.1NCERT Example 1 · p. 135

Coconut palm ki height

Level ground par ek coconut palm seedha khada hai. Point $P$ ground par hai, uske foot se 39 m door. $P$ se palm ke top ka angle of elevation hai $30^{\circ} .$ Palm ki height nikaalo. (Fig. 9.5)

Pehle khud try karo. Phir solution ek-ek step karke kholo.

2/3Rise $13.2 - 1.2 = 12$ m hai. Ladder $L$ hypotenuse hai aur $b$ horizontal side hai.

Fig. 9.6Example 9.2: ladder ko sirf deck se sill tak chadhna hai.

Road ke left mein ek fire engine khada hai aur right mein windows wali ek building hai. Ladder truck ke deck par lage hinge se uthti hai, jo road se 1.2 m upar hai, aur road se 13.2 m upar window sill tak jaati hai; dono heights dimension arrows se mark hain. Step 1 scene dikhata hai. Step 2 ladder L, hinge se wall tak dashed horizontal b aur 12 m ki dashed vertical se bane right triangle ko shade karta hai, aur hinge par 60 degree angle dikhata hai. Step 3 answers dikhata hai: L equals 8 root 3, lagbhag 13.86 m, aur b equals 4 root 3, lagbhag 6.93 m.

Example 9.2NCERT Example 2 · p. 136

Fire engine ki ladder

Ek fire engine building ke paas park hai. Uski ladder truck ke deck par hinge se judi hai, jo road se 1.2 m upar hai, aur use road se 13.2 m upar ek window sill tak pahunchna hai. Ladder ko horizontal ke saath $60^{\circ}$ banana hai. Ladder kitni lambi honi chahiye, aur hinge wall se kitni door hai? Use karo $\sqrt{3} \approx 1.732.$ (Fig. 9.6)

Pehle khud try karo. Phir solution ek-ek step karke kholo.

1/2Uski eye par angle of elevation $60^{\circ}$ hai; woh 11 m door hai aur uski eyes 1.6 m oonchi hain.

Fig. 9.7Example 9.3: right triangle Meera ke eye level par hai, ground par nahi.

Left mein chaar pairon par ek water tank khada hai aur right mein uske foot se 11 m door ek ladki khadi hai. Uski 1.6 m ki eye height dimension arrow se mark hai. Uski eye se tower tak ek dashed horizontal jaati hai aur line of sight 60 degrees par tank ke top tak uthti hai. Step 1 yeh givens dikhata hai. Step 2 uske eye level ke upar wale right triangle ko shade karta hai, left mein unknown side x mark karta hai, aur dikhata hai x equals 11 root 3, lagbhag 19.05 m, aur total height lagbhag 20.65 m.

Example 9.3NCERT Example 3 · p. 137

Eye height ko ginna

Meera ki eyes ground se 1.6 m upar hain. Woh apni colony ke water-tank tower ke foot se 11 m door khadi hai. Wahan se tank ke top ka angle of elevation hai $60^{\circ} .$ Tank ke top tak tower kitna ooncha hai? Use karo $\sqrt{3} \approx 1.732.$ (Fig. 9.7)

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Ek triangle

3 mein se 1

Flagpole ke foot se 25 m door ek point se, ground level par measure kiya gaya uske top ka angle of elevation hai

45^{\circ} .

Flagpole kitna ooncha hai?

Answer dekho

Answer 25 m

h = 25 \tan 45^{\circ} = 25 \times 1 = 25 m .

45^{\circ}

par height hamesha distance ke barabar hoti hai.

Do right triangles wale problems

Mushkil problems mein do angles diye hote hain. Tab figure mein do right triangles hote hain jo ek side share karte hain. Wahi shared side key hai. Har triangle se ek equation likho, phir jo side nahi chahiye use eliminate karo.

Neeche ke examples mein teen set-ups par dhyaan do.

Ek object doosre par khada hai, ek point se dekha gaya (Example 9.4).
Ek object ek line par do alag angles se dekha gaya (Example 9.5).
Ek oonche point se do angles of depression (Example 9.6 aur Example 9.7).

1/340 m oonchi cliff ke kinaare par ek lighthouse khada hai. Samundar mein ek boat hai.

Fig. 9.8Example 9.4: dono triangles share karte hain distance

d .

Right mein 40 m oonchi ek vertical cliff hai jiske kinaare par ek dhaariyon wala lighthouse hai; left mein samundar par ek boat B tair rahi hai. Step 1 scene dikhata hai. Step 2 neeche wala right triangle banata hai: boat se cliff ke foot tak sea-level distance d, vertical side ke roop mein cliff, aur 30 degrees par cliff ke top tak line of sight, saath mein result d equals 40 root 3, lagbhag 69.28 m. Step 3 lighthouse ke top tak 45 degrees par ek dashed line of sight jodta hai aur lighthouse ki height h mark karta hai, saath mein result h equals 40 root 3 minus 40, lagbhag 29.28 m.

Example 9.4NCERT Example 4 · p. 137

Cliff par lighthouse

40 m oonchi ek vertical cliff ke kinaare par ek lighthouse khada hai. Samundar mein ek boat $B$ hai. Boat se cliff ke top (lighthouse ke foot) ka angle of elevation hai $30^{\circ},$ aur top par lage lamp ka $45^{\circ} .$ Lighthouse kitna ooncha hai? Boat cliff ke foot se kitni door hai? Use karo $\sqrt{3} \approx 1.732.$ (Fig. 9.8)

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Sun ki altitude ground se dekha gaya uska angle of elevation hai. Sun jitna upar chadhta hai, har shadow utni chhoti hoti jaati hai. Agla example isi ko ek measurement mein badal deta hai.

2/3Sun aur upar, 60° par ho to shadow chhoti hoti hai.

Fig. 9.9Example 9.5: jaise-jaise Sun upar chadhta hai, shadow chhoti hoti jaati hai.

Level ground ke right mein ek flag-mast khada hai. Step 1 mein Sun right mein neeche 45 degrees par hai, aur ek ray mast ke top ke upar se guzarkar left mein lambi shadow ke tip tak jaati hai. Step 2 mein Sun upar 60 degrees par hai, aur ek dashed ray chhoti shadow ke tip tak jaati hai. Step 3 dono rays aur dono shadows ek saath dikhata hai, dono tips ke beech 6 m ka difference, chhoti shadow x aur height h, saath mein result h equals 3 times 3 plus root 3, lagbhag 14.20 m.

Example 9.5NCERT Example 5 · p. 138

Flag-mast ki shadows

Ek dhoop wali subah mandir ke aangan mein lage flag-mast ki shadow do baar measure ki jaati hai. Pehli baar Sun ki altitude hai $45^{\circ} .$ Baad mein yeh ho jaati hai $60^{\circ},$ aur shadow 6 m chhoti ho gayi hai. Mast ki height nikaalo. Use karo $\sqrt{3} \approx 1.732.$ (Fig. 9.9)

Pehle khud try karo. Phir solution ek-ek step karke kholo.

2/3Corner hai $P,$ aur wahan se $A$ tak ki line horizontal se $30^{\circ}$ neeche jhukti hai aur $B$ tak ki line $60^{\circ}$ neeche.

Fig. 9.10Example 9.6: chhat ke ek corner se do angles of depression.NCERT Fig. 9.9 · p. 139

Left mein ek ooncha apartment block hai aur right mein 10 m ooncha ek ghar hai, beech mein road hai. P block ki chhat ka corner hai, A ghar ka top aur B uska foot. Step 1 scene dikhata hai. Step 2 P se ek dashed horizontal aur A aur B tak neeche jaati do lines of sight jodta hai, aur P par angles of depression 30 aur 60 degrees mark karta hai. Step 3 block se A tak ek dashed horizontal right-angle mark ke saath jodta hai, A aur B par barabar alternate angles, road ki width d, aur results d equals 5 root 3, lagbhag 8.66 m, aur h equals 15 m dikhata hai.

Example 9.6NCERT Example 6 · p. 139

Road ke us paar

Ek apartment block ke saamne road ke us paar 10 m ooncha ek ghar khada hai. Block ki chhat ke corner $P$ se dekho. Ghar ka top $A$ horizontal se $30^{\circ}$ neeche hai, aur uska foot $B$ usse $60^{\circ}$ neeche. Road ki width aur block ki height nikaalo. (Fig. 9.10)

Pehle khud try karo. Phir solution ek-ek step karke kholo.

2/3Drone hai $D,$ aur wahan se post $A$ horizontal se $45^{\circ}$ neeche hai aur post $B$ $60^{\circ}$ neeche.

Fig. 9.11Example 9.7: AB ke dono hisse do alag triangles se aate hain.NCERT Fig. 9.10 · p. 140

Ek chhota drone D level path ke upar hover kar raha hai, left mein post A aur right mein post B hai. Drone se uske theek neeche wale point tak 60 m ki ek dashed vertical hai. Step 1 scene dikhata hai. Step 2 drone se guzarti ek dashed horizontal aur A aur B tak lines of sight jodta hai, saath mein 45 aur 60 degrees ke angles of depression. Step 3 A aur B par barabar alternate angles aur dono ground distances mark karta hai, left mein 60 m aur right mein 20 root 3 m, aur total AB lagbhag 94.64 m.

Example 9.7NCERT Example 7 · p. 140

Canal path ke upar drone

Ek survey drone ek seedhe, level canal path se 60 m upar hover kar raha hai. Path par do marker posts $A$ aur $B$ hain, drone ke opposite sides par. Drone se unke angles of depression hain $45^{\circ}$ aur $60^{\circ} .$ Posts ek doosre se kitni door hain? Use karo $\sqrt{3} \approx 1.732.$ (Fig. 9.11)

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Do triangles

3 mein se 1

Ek tower ke top se, ek hi taraf khadi do cars ke angles of depression hain

30^{\circ}

aur

60^{\circ} .

Kaunsi car tower ke zyada paas hai?

30^{\circ}

wali car

60^{\circ}

wali cardono barabar door hain

Answer dekho

Answer (B) $60^{\circ}$ wali car

Zyada steep line of sight foot ke zyada paas girti hai: distance hai

\frac{h}{\tan θ},

jo bade angle ke liye chhoti hoti hai.

Practice · 16 questions

Practice 9.1: answers ke saath

NumericalEasyNCERT Exercise 9.1, Q1
Ek vertical radio mast ke top se ground mein gade ek peg tak steel ka guy wire tana hua hai. Wire 28 m lamba hai aur ground ke saath $60^{\circ}$ banata hai. Mast kitna ooncha hai? Use karo $\sqrt{3} \approx 1.732.$
m
Solution dikhao
Answer $14 \sqrt{3} \approx 24.25 m$
$\sin 60^{\circ} = \frac{h}{28},$ isliye $h = 28 \times \frac{\sqrt{3}}{2} .$
$h = 14 \sqrt{3}$ $\approx 24.25 m .$
NumericalMediumNCERT Exercise 9.1, Q2
Toofan mein ek bamboo pole toot jaata hai. Upar ka hissa alag hue bina jhuk jaata hai, aur uska tip foot se 6 m door ground ko chhoota hai, ground ke saath $60^{\circ}$ banate hue. Tootne se pehle pole kitna ooncha tha? Use karo $\sqrt{3} \approx 1.732.$
m
Solution dikhao
Answer $12 + 6 \sqrt{3} \approx 22.39 m$
Khada hissa $a$ : angle hai $60^{\circ},$ jiske yeh opposite hai, aur 6 m adjacent hai. $a = 6 \tan 60^{\circ} = 6 \sqrt{3} .$
Toota hissa $b$ : hypotenuse. $\cos 60^{\circ} = \frac{6}{b},$ isliye $b = 12.$
Height $= a + b$ $= 12 + 6 \sqrt{3}$ $\approx 22.39 m .$
3
NumericalEasyNCERT Exercise 9.1, Q3
Ek godown mein do ramps hain.
1. Hand-carts ke liye ek ramp ground ke saath $30^{\circ}$ par 1.2 m oonche loading platform tak uthta hai. Yeh kitna lamba hai?
  m
  Solution dikhao
  $\sin 30^{\circ} = \frac{1.2}{l},$ isliye $l = 1.2 \times 2 = 2.4 m .$
2. Boriyon ke liye ek chute 2.4 m oonche loft se floor tak $60^{\circ}$ par neeche aata hai. Yeh kitna lamba hai? Use karo $\sqrt{3} \approx 1.732.$
  m
  Solution dikhao
  $\sin 60^{\circ} = \frac{2.4}{l},$ isliye $l = \frac{4.8}{\sqrt{3}}$ $= 1.6 \sqrt{3}$ $\approx 2.77 m .$
NumericalEasyNCERT Exercise 9.1, Q4
Brick-kiln ki chimney ke base se 20 m door khade Sunil ko uska top horizontal se $60^{\circ}$ upar dikhta hai (ground level par measure kiya gaya). Chimney kitni oonchi hai? Use karo $\sqrt{3} \approx 1.732.$
m
Solution dikhao
Answer $20 \sqrt{3} \approx 34.64 m$
$\tan 60^{\circ} = \frac{h}{20},$ isliye $h = 20 \sqrt{3}$ $\approx 34.64 m .$
NumericalEasyNCERT Exercise 9.1, Q5
Ek weather balloon ko ground par lage hook se bandhe seedhe cable ke zariye ground se 75 m upar roka gaya hai. Cable ground ke saath $60^{\circ}$ banata hai. Cable kitna lamba hai? Use karo $\sqrt{3} \approx 1.732.$
m
Solution dikhao
Answer $50 \sqrt{3} \approx 86.60 m$
$\sin 60^{\circ} = \frac{75}{l},$ isliye $l = \frac{150}{\sqrt{3}}$ $= 50 \sqrt{3}$ $\approx 86.60 m .$
NumericalMediumNCERT Exercise 9.1, Q6
Kavya 43.4 m oonche ek mandir ke tower ki taraf seedhe chalti hai. Jab woh shuru karti hai, to use tower ka top apni eyes se guzarne wali horizontal se $30^{\circ}$ upar dikhta hai; jab woh rukti hai, to $45^{\circ}$ upar. Uski eyes ground se 1.4 m upar hain. Woh kitna chali? Use karo $\sqrt{3} \approx 1.732.$
m
Solution dikhao
Answer $42 (\sqrt{3} - 1) \approx 30.74 m$
Uski eyes ke upar height $= 43.4 - 1.4 = 42 m .$
$30^{\circ}$ par: distance $= \frac{42}{\tan 30^{\circ}} = 42 \sqrt{3} .$ $45^{\circ}$ par: distance $= 42.$
Chali $= 42 \sqrt{3} - 42$ $= 42 (\sqrt{3} - 1)$ $\approx 30.74 m .$
NumericalMediumNCERT Exercise 9.1, Q7
Ek purane kile ke 18 m oonche bastion ke top se ek flag mast uthta hai. Level ground par khada Rahul mast ke foot tak $45^{\circ}$ upar aur uske tip tak $60^{\circ}$ upar measure karta hai. Mast kitna lamba hai? Use karo $\sqrt{3} \approx 1.732.$
m
Solution dikhao
Answer $18 (\sqrt{3} - 1) \approx 13.18 m$
$\tan 45^{\circ} = \frac{18}{d},$ isliye $d = 18.$
$\tan 60^{\circ} = \frac{18 + m}{18},$ isliye $18 + m = 18 \sqrt{3} .$
$m = 18 (\sqrt{3} - 1)$ $\approx 13.18 m .$
NumericalMediumNCERT Exercise 9.1, Q8
Ek town square mein pedestal par ek freedom fighter ki 2.4 m oonchi bronze statue lagi hai. Ground par ek jagah se measure karne par pedestal ka top horizontal se $30^{\circ}$ upar hai aur statue ka sir $45^{\circ}$ upar. Pedestal kitna ooncha hai? Use karo $\sqrt{3} \approx 1.732.$
m
Solution dikhao
Answer $1.2 (\sqrt{3} + 1) \approx 3.28 m$
$\tan 30^{\circ} = \frac{p}{d}$ se milta hai $d = \sqrt{3} p .$
$\tan 45^{\circ} = \frac{p + 2.4}{d}$ se milta hai $p + 2.4 = \sqrt{3} p .$
To $p (\sqrt{3} - 1) = 2.4,$ aur $p = \frac{2.4}{\sqrt{3} - 1}$ $= 1.2 (\sqrt{3} + 1)$ $\approx 3.28 m .$
NumericalMediumNCERT Exercise 9.1, Q9
Level ground par ek 54 m ooncha mobile tower aur ek hostel khade hain. Tower ke foot se ground level par measure karne par hostel ki chhat horizontal se $30^{\circ}$ upar hai. Hostel ke foot se tower ka top $60^{\circ}$ upar hai. Hostel kitna ooncha hai?
m
Solution dikhao
Answer $18 m$
Hostel ke foot se: $\tan 60^{\circ} = \frac{54}{d},$ isliye $d = \frac{54}{\sqrt{3}} = 18 \sqrt{3} .$
Tower ke foot se: hostel $= d \tan 30^{\circ}$ $= 18 \sqrt{3} \times \frac{1}{\sqrt{3}} = 18 m .$
NumericalMediumNCERT Exercise 9.1, Q10
Ek sports ground 48 m chauda hai, aur uske aar-paar jaati ek seedhi line ke dono ends par ek-ek floodlight tower hai. Dono towers barabar oonche hain. Unke beech usi line par khadi Divya ko ek tower ka top horizontal se $60^{\circ}$ upar aur doosre ka $30^{\circ}$ upar dikhta hai (uski height ignore karo). Towers kitne oonche hain, aur woh har ek se kitni door hai? Use karo $\sqrt{3} \approx 1.732.$
Solution dikhao
AnswerTowers $12 \sqrt{3} \approx 20.78 m;$ woh unse 12 m aur 36 m door hai
$h = x \tan 60^{\circ} = \sqrt{3} x$ aur $h = (48 - x) \tan 30^{\circ} = \frac{48 - x}{\sqrt{3}} .$
To $3 x = 48 - x,$ jisse milta hai $x = 12.$ Doosri distance hai $36 m .$
$h = 12 \sqrt{3}$ $\approx 20.78 m .$
NumericalHardNCERT Exercise 9.1, Q11
Neha ek canal ke kinaare khadi hai, door wale kinaare par lage clock tower ke theek saamne, aur use uska top horizontal se $60^{\circ}$ upar dikhta hai. Woh paani se, tower se door, seedhe 10 m peeche chalti hai, aur ab top horizontal se $45^{\circ}$ upar dikhta hai. Uski height ignore karke batao, tower kitna ooncha hai aur canal kitni chaudi hai? Use karo $\sqrt{3} \approx 1.732.$
Solution dikhao
AnswerTower $5 (3 + \sqrt{3}) \approx 23.66 m;$ canal $\approx 13.66 m$ chaudi
Kinaare par: $h = w \tan 60^{\circ}$ $= \sqrt{3} w .$ Peeche hatne ke baad: $h = (w + 10) \tan 45^{\circ}$ $= w + 10.$
To $w (\sqrt{3} - 1) = 10,$ aur $w = 5 (\sqrt{3} + 1)$ $\approx 13.66 m .$
$h = w + 10$ $\approx 23.66 m .$
NumericalMediumNCERT Exercise 9.1, Q12
Ground se 15 m oonchi hostel terrace par khada Farhan ek mobile tower ke top ko horizontal se $60^{\circ}$ upar dekhta hai, aur uske foot ko $30^{\circ}$ neeche. Tower kitna ooncha hai?
m
Solution dikhao
Answer $60 m$
Foot: alternate angles se $\tan 30^{\circ} = \frac{15}{d},$ isliye $d = 15 \sqrt{3} .$
Top: terrace ke upar wala hissa $d \tan 60^{\circ}$ $= 15 \sqrt{3} \times \sqrt{3} = 45.$
Tower $= 15 + 45 = 60 m .$
NumericalMediumNCERT Exercise 9.1, Q13
Paani se 90 m oonche ek viewpoint se dekhne par ek jheel mein do boats ek seedhi line mein, ek ke peeche doosri, dikhti hain. Paas wali boat horizontal se $60^{\circ}$ neeche hai aur door wali $45^{\circ}$ neeche. Boats ek doosre se kitni door hain? Use karo $\sqrt{3} \approx 1.732.$
m
Solution dikhao
Answer $90 - 30 \sqrt{3} \approx 38.04 m$
Paas wali boat: $\frac{90}{\tan 60^{\circ}} = \frac{90}{\sqrt{3}} = 30 \sqrt{3} .$ Door wali boat: $\frac{90}{\tan 45^{\circ}} = 90.$
Gap $= 90 - 30 \sqrt{3}$ $\approx 90 - 51.96$ $= 38.04 m .$
NumericalMediumNCERT Exercise 9.1, Q14
Ek drone playground se 61.5 m upar level udta hai, Aman se door jaate hue. Jab woh pehli baar upar dekhta hai, to drone tak uski line of sight horizontal se $60^{\circ}$ upar hai; thodi der baad sirf $45^{\circ} .$ Uski eyes ground se 1.5 m upar hain. Dono baar dekhne ke beech drone kitna uda? Use karo $\sqrt{3} \approx 1.732.$
m
Solution dikhao
Answer $60 - 20 \sqrt{3} \approx 25.36 m$
Uski eyes ke upar height $= 61.5 - 1.5 = 60 m .$
Horizontal distances: $\frac{60}{\sqrt{3}} = 20 \sqrt{3}$ jab angle ho $60^{\circ},$ aur $60$ jab angle ho $45^{\circ} .$
Tay ki gayi distance $= 60 - 20 \sqrt{3}$ $\approx 25.36 m .$
NumericalHardNCERT Exercise 9.1, Q15
Watchtower ke top par khadi ek guard ek cyclist ko dekh rahi hai jo tower ke foot tak jaane wali seedhi road par ek steady speed se chal raha hai. Jab woh pehli baar dekhti hai, to cyclist uski horizontal se $30^{\circ}$ neeche hai; 10 seconds baad $45^{\circ}$ neeche. Cyclist ko tower ke foot tak pahunchne mein aur kitne seconds lagenge? Use karo $\sqrt{3} \approx 1.732.$
s
Solution dikhao
Answer $5 (\sqrt{3} + 1) \approx 13.66 s$
Maan lo tower $h$ m ooncha hai. Distances hain $h \sqrt{3}$ jab angle $30^{\circ}$ ho, aur $h$ jab angle ho $45^{\circ} .$
10 s mein cyclist cover karta hai $h (\sqrt{3} - 1),$ isliye aakhri $h$ metres mein $\frac{10 h}{h (\sqrt{3} - 1)}$ seconds lagte hain.
Time $= \frac{10}{\sqrt{3} - 1}$ $= 5 (\sqrt{3} + 1)$ $\approx 13.66 s .$
Quick answerEasy
9 m ooncha ek pole $3 \sqrt{3}$ m lambi shadow banata hai. Sun ki altitude hai
$30^{\circ}$ $45^{\circ}$ $60^{\circ}$ $90^{\circ}$
Answer dekho
Answer (C) $60^{\circ}$
$\tan θ = \frac{9}{3 \sqrt{3}} = \sqrt{3},$ isliye $θ = 60^{\circ} .$ $30^{\circ}$ ke liye shadow ko height ka $\sqrt{3}$ guna hona padta.

9.2 Summary

Recap

Summary aur key points

Line of sight observer ki eye ko us point se jodti hai jise dekha ja raha hai (Line of sight aur angle of elevation).
Jab point eye level se upar ho, to angle of elevation eye se guzarne wali horizontal se upar ki taraf measure hota hai (Fig. 9.1).
Jab point eye level se neeche ho, to angle of depression eye se guzarne wali horizontal se neeche ki taraf measure hota hai (Angle of depression).
$A$ se $B$ ka angle of depression, $B$ se $A$ ke angle of elevation ke barabar hota hai: yeh alternate angles hain (key idea).
Angle of elevation se height: $H = e + d \tan θ,$ jahan $e$ eye height hai ((9.1)).
Height aur horizontal distance ke liye $\tan$ use karo; jab ladder, rassi ya dor (hypotenuse) involved ho, to $\sin$ ya $\cos$ use karo (Table 9.2).
Do angles hon to woh side dhoondho jo do right triangles share karte hain, aur har triangle ke liye ek equation likho (Do right triangles wale problems).
$40 \sqrt{3}$ jaise surds ko aakhir tak rakho, phir $\sqrt{3} \approx 1.732$ use karo (Table 9.1).

1/3Height aur horizontal distance: opposite aur adjacent, isliye use karo $\tan θ .$

Fig. 9.12Kaunsa ratio? Dekho tumhe kaunsi sides pata hain aur kaunsi chahiye.

Teen steps mein ek recap. Step 1 ek right triangle dikhata hai jismein height aur horizontal distance highlight hain, neeche left mein angle theta hai aur likha hai tan lo, opposite aur adjacent. Step 2 wahi triangle dikhata hai jismein slanting side highlight hai aur us par ladder, rassi, dor likha hai, saath mein sin ya cos lo. Step 3 do observers se ek oonchi vertical side ke top tak bane do right triangles dikhata hai; yeh side shared side ke roop mein highlight hai, saath mein angles alpha aur beta aur likha hai ek-ek equation.

Flash cards · 5

Jaldi yaad karo

1 / 5

Board-exam practice

Yeh questions CBSE board paper ke formats follow karte hain: multiple choice, assertion–reason, short aur long answers, aur ek case study.

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 9.1, Q1	Practice Q1
NCERT Exercise 9.1, Q2	Practice Q2
NCERT Exercise 9.1, Q3	Practice Q3
NCERT Exercise 9.1, Q4	Practice Q4
NCERT Exercise 9.1, Q5	Practice Q5
NCERT Exercise 9.1, Q6	Practice Q6
NCERT Exercise 9.1, Q7	Practice Q7
NCERT Exercise 9.1, Q8	Practice Q8
NCERT Exercise 9.1, Q9	Practice Q9
NCERT Exercise 9.1, Q10	Practice Q10
NCERT Exercise 9.1, Q11	Practice Q11
NCERT Exercise 9.1, Q12	Practice Q12
NCERT Exercise 9.1, Q13	Practice Q13
NCERT Exercise 9.1, Q14	Practice Q14
NCERT Exercise 9.1, Q15	Practice Q15

Quick check: MCQs

Elevation ya depression?4 questions
Ratio chuno3 questions
Ek triangle3 questions
Do triangles3 questions

Glossary · 8

Key terms (English + Hindi)

alternate anglesएकांतर कोणekantar kon: Do lines ko kaatne wali ek line ke opposite sides par, un do lines ke beech bane do angles. Jab do lines parallel hon, to yeh barabar hote hain.Chapter mein pehli baar yahan aaya →
altitude of the Sunसूर्य का उन्नतांशsurya ka unnatansh: Ground se dekha gaya Sun ka angle of elevation. Sun jitna ooncha, shadows utni chhoti.Chapter mein pehli baar yahan aaya →
angle of depressionअवनमन कोणavanaman kon: Jab dekha gaya point eye level se neeche ho, tab line of sight aur horizontal ke beech ka angle.Chapter mein pehli baar yahan aaya →
angle of elevationउन्नयन कोणunnayan kon: Jab dekha gaya point eye level se upar ho, tab line of sight aur horizontal ke beech ka angle.Chapter mein pehli baar yahan aaya →

clinometer: Angles of elevation ya depression measure karne ka ek simple instrument, jaise straw aur latakte weight wala protractor.Chapter mein pehli baar yahan aaya →

horizontalक्षैतिजkshaitij: Observer ki eye se guzarne wali level line, jo level ground ke parallel hoti hai. Angles of elevation aur depression isi se measure hote hain.Chapter mein pehli baar yahan aaya →

line of sightदृष्टि रेखाdrishti rekha: Observer ki eye se dekhe ja rahe point tak ki seedhi line.Chapter mein pehli baar yahan aaya →

trigonometric ratioत्रिकोणमितीय अनुपातtrikonmitiya anupat: Right triangle ki do sides ka ratio, jo uske kisi ek acute angle se juda ho, jaise $\tan θ = \frac{opposite}{adjacent} .$ Chapter mein pehli baar yahan aaya →

Some Applications of Trigonometry: Class 10 Maths Chapter 9 Hinglish mein

Is chapter mein aap kya seekhenge

9.1 Heights and Distances

Line of sight aur angle of elevation

Angle of depression

Heights problem ka plan banana

Ek right triangle wale problems

Do right triangles wale problems

Practice 9.1: answers ke saath

9.2 Summary

Summary aur key points

Board-exam practice

Practice questions answers ke saath (NCERT “Board-exam practice”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye