Introduction to Trigonometry: Class 10 Maths Chapter 8 Hinglish mein

Right triangle ke chhah ratios, special angles par unki values, aur unhe jodne wali identities

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

Published 12 Oct 2026 · Updated 12 Oct 2026

75 min
32 practice questions
20 MCQs

Is chapter mein aap kya seekhenge

Right triangle ke kisi bhi acute angle ke respect mein opposite side, adjacent side aur hypotenuse pehchaano.
Acute angle ke sin, cos, tan, cosec, sec aur cot define karo, aur samjhao ki unki values sirf angle par depend karti hain, triangle ke size par nahi.
Jab ek trigonometric ratio diya ho, to Pythagoras theorem se us angle ke saare chhah ratios nikaalo.
Simple triangles se 30°, 45° aur 60° ke ratios derive karo, 0° aur 90° par values samjhao, aur batao ki wahan kaunse ratios defined nahi hain.
Special values se expressions evaluate karo aur right triangles ki unknown sides aur angles nikaalo.
sin²A + cos²A = 1, 1 + tan²A = sec²A aur 1 + cot²A = cosec²A prove karo, batao ki har ek kahan valid hai, aur unse doosri identities prove karo.

NCERT Mathematics Class 10, Chapter 8, pp. 113–132 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

8.1 Introduction

Aas-paas dekho, tumhe har jagah right triangles milenge, wahan bhi jahan kisi ne unhe draw nahi kiya: tirchi khadi ladder mein, patang ki dor mein, chhat tak jaati tumhari line of sight mein. Inka sabse famous fact tum pehle se jaante ho, Pythagoras theorem. Yeh chapter tumhe ek doosra, aur zyada powerful tool deta hai.

Makar Sankranti par patang. Meera ne lambi dor chhodi hai, jo hawa se tight khinchi hui hai. Dor, patang se neeche girti ek vertical line aur uske haath se nikalti ek level line milkar right triangle banate hain. Kya woh kahin chadhe bina bata sakti hai ki patang kitni oonchi hai?
Painter ki ladder. Ek ladder ghar ki deewar se tiki hai. Ladder, deewar aur zameen ek right triangle banate hain. Agar tumhe ladder ki length pata hai aur yeh ki woh kitni tirchi hai, to kya tum bata sakte ho ki woh deewar par kitni oonchai tak pahunchti hai?
Mobile tower. Tum ek tower ke base se 40 m door khade ho aur uske top ko dekhte ho. Tumhari line of sight, tower aur tumhari aankh se nikalti level line ek right triangle banate hain. Kya tum apni jagah chhode bina tower ki height nikaal sakte ho?

3/3? se mark ki gayi heights ko seedhe naapna mushkil hai. Trigonometry unhe un lengths aur angles se nikaalti hai jo naapna aasaan hai.

Fig. 8.1Roz ke scenes mein chhupe right triangles.NCERT Figs. 8.1–8.3 · p. 113

Level zameen par roz ke teen scenes. Left mein ek bachcha seedhi dor par patang uda raha hai; beech mein ek ladder ghar ki deewar se tiki hai; right mein ek insaan ek oonche lattice mobile tower se 40 m door khada hai aur uske top ko dekh raha hai. Pehle step mein sirf scenes bane hain. Doosre step mein har scene mein ek shaded right triangle aata hai, jismein right-angle mark hai aur neeche ke corner par acute angle mark hai: dor, vertical drop aur level line; ladder, deewar aur zameen; line of sight, tower aur level line. Teesre step mein unknown heights question marks se mark hoti hain aur 40 m distance par label lagta hai.

Har case mein, jis length tak pahunchna mushkil hai woh ek right triangle ki side hai, jiske baaki parts naapna aasaan hai. In parts ko jodne wale tools trigonometry ke hain. Yeh naam Greek trigōnon (triangle, literally "teen corners") aur metron (naap) ko jodta hai. To trigonometry ka literal matlab hai triangle ka naap: yeh triangle ki sides ki lengths ko uske angles ke size se jodti hai.

Aaj har GPS receiver, zameen ka plot naapne wala har surveyor aur bridge design karne wala har engineer trigonometry par depend karta hai. Iski jadein bahut purani hain. Egyptian aur Babylonian builders aur aasmaan dekhne waalon ne iske ideas use kiye, aur Greek aur Indian astronomers ne isse tables banaye taaki andaaza laga sakein ki Moon aur planets kitni door hain.

Timeline

Zamaane ke saath trigonometry

c. 1800 BCE
Ek Babylonian mitti ki tablet
Jis tablet ko aaj Plimpton 322 kehte hain, usmein aise numbers likhe hain jo whole-number sides wale right triangles se match karte hain.
Doosri sadi BCE
Hipparchus ki chord table
Greek astronomer Hipparchus ne aasmaan ko study karne ke liye circle ki chords ki lengths ki list banayi.
499 CE
Āryabhaṭīya
Āryabhaṭa ne half-chords ki ek table di, jo aaj ke sine ki purvaj hai.
1600s
Chhote naam
England ke writers ne short forms "sin" aur "cos" use karna shuru kiya.

Yeh chapter teen sawaalon par chalta hai. Right triangle ke kisi angle ke kaunse ratios of sides hote hain, aur triangle ka size kyun matter nahi karta? $30^{\circ}$ , $45^{\circ}$ aur $60^{\circ}$ par in ratios ki exact values kya hain, aur extremes $0^{\circ}$ aur $90^{\circ}$ par kya hota hai? Aur ratios ke beech kaunse relations, jinhe identities kehte hain, har angle ke liye sahi hote hain?

Quick check

2 mein se 1

Ladder wale scene mein right triangle ka kaunsa angle

90^{\circ}

hai?

ladder aur zameen ke beech ka angleladder aur deewar ke beech ka angledeewar aur zameen ke beech ka angleladder ke top par ka angle

Answer dekho

Answer (C) deewar aur zameen ke beech ka angle

Deewar level zameen se seedhi upar khadi hai, isliye right angle wahan hai jahan deewar zameen se milti hai. Ladder dono ke saath acute angles banati hai.

8.2 Trigonometric Ratios

Ek right triangle $A B C$ lo jiska right angle $B$ par hai. Uske baaki do angles, $∠ A$ aur $∠ C$ , acute hain, aur dono milkar $90^{\circ}$ banate hain. Apna dhyaan $∠ A$ par rakho, jise simply $A$ likhte hain. Ab har side ko ek naam milta hai jo batata hai ki woh $A$ ke respect mein kahan hai.

Hypotenuse $A C$ right angle ke saamne hai. Yeh sabse lambi side hai, aur iska naam kabhi nahi badalta.
$A$ ki opposite side $B C$ hai. Yahi akeli side hai jo $A$ ko touch nahi karti.
$A$ ki adjacent side $A B$ hai. Hypotenuse ke saath milkar yeh angle $A$ banati hai.

1/2 $A$ se: $B C$ opposite hai, $A B$ adjacent hai aur $A C$ hypotenuse hai.

Fig. 8.2Opposite aur adjacent is par depend karte hain ki tum kis angle se dekh rahe ho.NCERT Figs. 8.4, 8.5 · p. 114

Triangle ABC ka right angle B par hai, AB neeche ki taraf hai aur BC seedhi khadi hai. Pehle step mein angle A shaded hai; BC par opposite A, AB par adjacent to A aur AC par hypotenuse likha hai. Doosre step mein uski jagah angle C shaded hai; ab AB par opposite C aur BC par adjacent to C likha hai, jabki AC ab bhi hypotenuse hai. Dono steps mein opposite side orange aur adjacent side blue hai, isliye dono chhoti sides naam ke saath-saath colours bhi badalti hain.

Agar tum $∠ C$ se dekho, to dono chhoti sides ke naam badal jaate hain: $A B$ , $C$ ki opposite side ban jaati hai, aur $B C$ , $C$ ki adjacent side. To "opposite" aur "adjacent" ka matlab tabhi banta hai jab tum batao ki tum kaunse angle se dekh rahe ho.

Chhah ratios

Right triangle ki do sides ka ratio, jo uske kisi acute angle ke respect mein liya jaaye, us angle ka trigonometric ratio kehlaata hai. Teen sides ko chhah tarah se pair kiya ja sakta hai, aur har tarah ka ek naam hai.

Table 8.1

△ A B C

mein, jo

B

par right-angled hai,

∠ A

ke chhah trigonometric ratiosNCERT p. 115

Ratio (naam)	Kaunsi sides	△ABC mein
$\sin A$ (sine)	opposite ÷ hypotenuse	$\frac{B C}{A C}$
$\cos A$ (cosine)	adjacent ÷ hypotenuse	$\frac{A B}{A C}$
$\tan A$ (tangent)	opposite ÷ adjacent	$\frac{B C}{A B}$
$cosec A$ (cosecant)	hypotenuse ÷ opposite	$\frac{A C}{B C}$
$\sec A$ (secant)	hypotenuse ÷ adjacent	$\frac{A C}{A B}$
$\cot A$ (cotangent)	adjacent ÷ opposite	$\frac{A B}{B C}$

Sine, cosine aur tangent ko tum sabse zyada use karoge. Table ke aakhri teen ratios pehle teen ke reciprocals hain:

\begin{aligned} cosec A & = \frac{1}{\sin A} \\ \sec A & = \frac{1}{\cos A} \\ \cot A & = \frac{1}{\tan A} \end{aligned}

$\frac{B C}{A B}$ ke upar aur neeche dono ko hypotenuse $A C$ se divide karne par tangent, sine aur cosine se jud jaata hai:

\tan A = \frac{\sin A}{\cos A}

(8.1)

\cot A = \frac{\cos A}{\sin A}

(8.2)

Symbols ke baare mein ek warning. $3 x$ mein tum 3 ko alag kar sakte ho, lekin $\sin A$ mein "sin" ko $A$ se alag nahi kar sakte: poora symbol ek number ka naam hai, angle $A$ ka sine. Akela "sin" likhna aisa hai jaise square-root ka sign likh do aur uske neeche kuch na likho. Yahi baat $\cos A$ , $\tan A$ aur baaki sab par laagu hoti hai.

History se

"Sine" word kahan se aaya

Lagbhag 499 CE mein Āryabhaṭa ne apni astronomy mein circle ki half-chord use ki. Unhone use ardha-jyā (अर्धज्या, "aadhi dhanush ki dori") kaha, jo baad mein chhota hokar jyā ya jīvā ban gaya. Unka kaam aage badhane wale Arab scholars ne yeh word jība likha. Jab barahvi sadi mein unki kitaabon ka Latin mein translation hua, to is word ko ek milte-julte Arabic word samajh liya gaya jiska matlab tha fold ya pocket, aur yeh Latin sinus ban gaya. Sinus se humein "sine" mila.

Indian astronomers ne triangle ke doosre acute angle ka sine bhi koṭi-jyā naam se table mein likha; wahi hamara cosine hai. Short forms kai sadiyon baad aaye. English astronomer Edmund Gunter (1581–1626) ko short form "sin" ka credit diya jaata hai, aur "cos" pehli baar 1674 mein Sir Jonas Moore ke kaam mein dikha.

Ratios angle par depend karte hain, size par nahi

Agar hum usi angle ke saath bada triangle banayein, to kya angle ka sine badalta hai? Fig. 8.3 dekho. Points $D$ , $E$ aur $C$ , $∠ A$ ki ek arm par hain, aur perpendiculars $D F$ , $E G$ aur $C B$ doosri arm par girte hain.

4/4Har ratio lagbhag 0.47 hai: yeh angle $A$ ka hai, kisi ek triangle ka nahi.

Fig. 8.3Angle same, to ratio bhi same:

\frac{8}{17} = \frac{16}{34} = \frac{24}{51} .

NCERT Fig. 8.6 · p. 116

Ek angle A jiski tirchi arm par points D, E aur C hain, aur unse horizontal arm par perpendiculars DF, EG aur CB giraye gaye hain, jo right triangles ADF, AEG aur ACB banate hain. Unki opposite sides 8, 16 aur 24 units hain aur hypotenuses 17, 34 aur 51 units. Pehla step chhota triangle shade karta hai aur DF ÷ AD = 8 ÷ 17 dikhata hai. Doosra step beech wala triangle shade karta hai aur EG ÷ AE = 16 ÷ 34 jodta hai. Teesra step bada triangle shade karta hai aur CB ÷ AC = 24 ÷ 51 jodta hai. Aakhri step teenon triangles dikhata hai aur note karta hai ki har ratio lagbhag 0.47 hai.

Triangles $A D F$ , $A E G$ aur $A C B$ angle $A$ share karte hain, aur har ek mein ek right angle hai. To Chapter 6 ke AA similarity criterion se yeh similar hain, aur inki matching sides proportion mein hain. Isse milta hai

\frac{D F}{A D} = \frac{E G}{A E} = \frac{C B}{A C}

In mein se har fraction $A$ ki opposite side ko hypotenuse se divide karke bana hai, isliye har ek $\sin A$ hai. Isi tarah $\frac{A F}{A D} = \frac{A G}{A E} = \frac{A B}{A C},$ isliye teenon triangles mein $\cos A$ same hai, aur baaki har ratio bhi. Trigonometric ratio ki value sirf angle par depend karti hai, is par nahi ki triangle kitna bada hai. Isiliye " $35^{\circ}$ ka sine" kehne ka koi matlab banta hai.

Try karke dekho

35° ka sine khud naapo

Tumhe chahiye

ek ruler aur ek protractor
ek set square
ek nukeeli pencil

Pehle guess karo

Tum

35^{\circ}

angle ki ek arm par corner se 5 cm, 8 cm aur 10 cm par points mark karoge aur perpendiculars giraoge. Kaunsa point opposite ÷ hypotenuse ki sabse badi value dega?

5 cm wala point8 cm wala point10 cm wala pointsab lagbhag same value dete hain

Ek ray $A X$ draw karo aur protractor se ek angle $∠ X A Y = 35^{\circ}$ banao.
$A Y$ par $A$ se 5 cm, 8 cm aur 10 cm par points mark karo.
Set square se har point se $A X$ par perpendicular girao.
Har perpendicular naapo aur use uske point ki $A$ se distance se divide karo.

Table 8.2

Hypotenuse	Opposite	Ratio
5 cm
8 cm
10 cm

Tumhari entries apne aap save hoti hain.

InteractiveRight-triangle explorer

Ek right triangle ABC jiska right angle B par hai. Tum corner C ko drag kar sakte ho, ya do sliders use kar sakte ho: ek angle A ko 0° se 90° tak set karta hai, doosra sirf hypotenuse badalta hai, 3 cm se 10 cm tak. Opposite side, adjacent side aur hypotenuse alag colours mein hain aur un par unki lengths likhi hain. Ek table saare chhah ratios dikhata hai, unki sides aur values ke saath, aur 30°, 45° aur 60° par 1/2 ya √3/2 jaise exact forms, aur jahan 0° ya 90° par ratio zero se divide hota, wahan "not defined" likha hai. Buttons view ko angle C par le jaate hain, special angles par jump karte hain, aur Play angle ko 0° se 90° tak ghumata hai. Neeche ek line check karti hai ki sin² + cos² = 1.

Fig. 8.4Angle badalne ke liye

C

ko drag karo, ya sirf size badlo aur dekho ki ratios wahi rehte hain.

Quick check

4 mein se 1

△ P Q R

mein, jo

Q

par right-angled hai, kaunsa fraction

\tan R

hai?

\frac{P Q}{Q R}

\frac{Q R}{P Q}

\frac{P Q}{P R}

\frac{Q R}{P R}

Answer dekho

Answer (A) $\frac{P Q}{Q R}$

R

se opposite side

P Q

hai aur adjacent side

Q R

, isliye

\tan R = \frac{P Q}{Q R}

. Lalchane wala

\frac{Q R}{P Q}

asal mein

\tan P

hai: yeh galat angle se dekhta hai.

Ek ratio se saare chhah tak

Acute angle ka ek ratio pata ho to saare chhah nikaalne ke liye kaafi hai. Maan lo tumhe sirf itna bataya gaya ki $\cos A = \frac{2}{3}$ , jahan $△ A B C$ mein $∠ B = 90^{\circ}$ .

Koi bhi right triangle draw karo jismein adjacent side aur hypotenuse $2 : 3$ ke ratio mein hon. Kyunki size matter nahi karta, kisi positive number $k$ ke liye unhe $A B = 2 k$ aur $A C = 3 k$ maan lo. Missing side Pythagoras theorem se milti hai:

\begin{aligned} B C^{2} & = A C^{2} - A B^{2} \\ = 9 k^{2} - 4 k^{2} \\ = 5 k^{2} \end{aligned}

To $B C = \sqrt{5} k$ . Hum sirf positive root lete hain, kyunki length negative nahi ho sakti. Ab har ratio definitions se nikal aata hai: $\sin A = \frac{\sqrt{5}}{3}$ , $\tan A = \frac{\sqrt{5}}{2}$ , $cosec A = \frac{3}{\sqrt{5}}$ , $\sec A = \frac{3}{2}$ aur $\cot A = \frac{2}{\sqrt{5}}$ . Dhyaan do ki $k$ har baar cancel ho jaata hai, jaisa hona chahiye.

Example 8.1NCERT Example 1 · p. 118

Har 35 cm par 12 cm uthne wala ramp

School gate par ek ramp zameen ke saath har 35 cm chalne par 12 cm upar uthta hai. To zameen ke saath uske angle $A$ ke liye $\tan A = \frac{12}{35}$ hai. $A$ ke baaki paanch trigonometric ratios nikaalo.

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

Example 8.2NCERT Example 2 · p. 118

Barabar tangents matlab barabar angles

Do acute angles, $∠ A$ aur $∠ P$ , ka tangent same hai. Prove karo ki angles barabar hain.

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

Yahi baat har ratio ke liye sach hai: agar do acute angles ke sines barabar hon, ya cosines barabar hon, to angles barabar hote hain. Cosine wala case tum Practice 8.1 mein prove karoge.

Example 8.3NCERT Example 3 · p. 119

Difference of squares se teesri side

Right triangle $P Q R$ ( $∠ Q = 90^{\circ}$ ) ka hypotenuse $P R$ 65 units lamba hai aur side $Q R$ 33 units. Maan lo $θ$ , $R$ par ka angle hai. (i) $\sin^{2} θ + \cos^{2} θ$ aur (ii) $\sin^{2} θ - \cos^{2} θ$ nikaalo.

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

Part (i) ka answer bilkul 1 aaya. Yeh ittefaq nahi hai: Section 8.4 mein tum prove karoge ki har angle $θ$ ke liye $\sin^{2} θ + \cos^{2} θ = 1$ .

Example 8.4NCERT Example 4 · p. 120

Aadha moda hua square napkin

Ek square napkin ko diagonal par moda jaata hai, jisse $△ A B C$ banta hai jismein $∠ B = 90^{\circ}$ aur $A B = B C$ . Dikhao ki $\sin A = \cos A$ , aur $\sin^{2} A + 2 \sin A \cos A$ ki value nikaalo.

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

Example 8.5NCERT Example 5 · p. 120

Ek side aur ek difference

Right triangle $P Q R$ mein, jismein $∠ Q = 90^{\circ}$ , side $P Q$ 28 cm hai aur hypotenuse $Q R$ se 8 cm lamba hai. $\sin R$ aur $\cos R$ nikaalo.

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

Quick check

3 mein se 1

Agar acute angle

A

ke liye

\sin A = \frac{5}{13}

hai, to

\cot A

kya hai?

\frac{5}{12}

\frac{12}{5}

\frac{13}{12}

\frac{12}{13}

Answer dekho

Answer (B) $\frac{12}{5}$

Opposite

= 5 k

aur hypotenuse

= 13 k

, isliye adjacent side

12 k

hai. Tab

\cot A

= adjacent ÷ opposite

= \frac{12}{5}

. Answer

\frac{5}{12}

\tan A

hai, yaani wahi fraction ulta.

8.3 Kuch Specific Angles ke Trigonometric Ratios

Board papers mein kuch angles baaki sab se kahin zyada aate hain: $30^{\circ}$ , $45^{\circ}$ aur $60^{\circ}$ , aur saath mein do extremes $0^{\circ}$ aur $90^{\circ}$ . Pehle teen aisi shapes se aate hain jinhe tum compass aur ruler se draw kar sakte ho, aur unke ratios ki exact values hain jo tumhe zubaani yaad honi chahiye.

45° ke ratios

Ek square ko diagonal par kaato (Fig. 8.5). Har aadha hissa ek right triangle hai jiski do chhoti sides barabar hain, isliye uske do acute angles bhi barabar hain. Kyunki unka sum $90^{\circ}$ hai, har ek $45^{\circ}$ hai.

3/3Diagonal $\sqrt{2} a$ hai, isliye $\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$ aur $\tan 45^{\circ} = 1$ .

Fig. 8.5Aadha square

45^{\circ}

ke ratios deta hai.NCERT Fig. 8.14 · p. 122

Side a wala ek square jismein dashed diagonal AC hai. Pehle step mein poora square dikhta hai. Doosre step mein aadha triangle ABC, jo B par right-angled hai, shaded hota hai aur uske dono acute angles par 45° mark hota hai, jabki doosra aadha halka pad jaata hai. Teesre step mein diagonal par root 2 times a likha jaata hai, aur triangle ke paas ratios sin 45° = 1/√2, cos 45° = 1/√2 aur tan 45° = 1 dikhte hain.

Derivation45° ke ratios

Maan lo aadha square $△ A B C$ hai jiska right angle $B$ par hai, taaki $∠ A = ∠ C = 45^{\circ}$ . Barabar angles ke saamne ki sides barabar hoti hain, isliye $A B = B C$ ; dono ko $a$ kaho. Pythagoras theorem se,

\begin{aligned} A C^{2} & = a^{2} + a^{2} \\ = 2 a^{2} \\ A C & = \sqrt{2} a \end{aligned}

Ab definitions se ratios padh lo; $a$ har baar cancel ho jaata hai:

Ratio	Sides	Value
$\sin 45^{\circ}$	$\frac{B C}{A C} = \frac{a}{\sqrt{2} a}$	$\frac{1}{\sqrt{2}}$
$\cos 45^{\circ}$	$\frac{A B}{A C} = \frac{a}{\sqrt{2} a}$	$\frac{1}{\sqrt{2}}$
$\tan 45^{\circ}$	$\frac{B C}{A B} = \frac{a}{a}$	$1$

Inke reciprocals $cosec 45^{\circ} = \sqrt{2}$ , $\sec 45^{\circ} = \sqrt{2}$ aur $\cot 45^{\circ} = 1$ hain.

30° aur 60° ke ratios

In do angles ke liye side $2 a$ wale ek equilateral triangle $P Q R$ se shuru karo, jismein har angle $60^{\circ}$ hai. $P$ se $Q R$ par perpendicular $P S$ girao (Fig. 8.6).

4/5 $P$ se: opposite $a$ , adjacent $\sqrt{3} a$ , hypotenuse $2 a$ .

Fig. 8.6Aadha equilateral triangle

30^{\circ}

aur

60^{\circ}

ke ratios deta hai.NCERT Fig. 8.15 · p. 122

Ek equilateral triangle PQR jiski sides 2a hain aur saare angles 60°. Doosre step mein altitude PS, QR ke midpoint S tak draw hota hai, isliye QS = SR = a, aur upar ka angle do 30° angles mein bant jaata hai. Teesre step mein aadha triangle PQS shaded hota hai, jismein PQ = 2a, QS = a aur altitude PS = root 3 times a. Chauthe step mein P par 30° angle mark hota hai aur ratios sin 30° = 1/2, cos 30° = √3/2 aur tan 30° = 1/√3 dikhte hain. Paanchve step mein Q par 60° angle mark hota hai aur ratios sin 60° = √3/2, cos 60° = 1/2 aur tan 60° = √3 dikhte hain.

Derivation30° aur 60° ke ratios

Triangles $P Q S$ aur $P R S$ , $S$ par right-angled hain. Unke hypotenuses $P Q$ aur $P R$ barabar hain, aur dono side $P S$ share karte hain, isliye RHS se yeh congruent hain. Isliye $Q S = S R = a$ aur $∠ Q P S = ∠ R P S = 30^{\circ}$ .

Ab right triangle $P Q S$ mein kaam karo, jahan $∠ Q = 60^{\circ}$ , $∠ Q P S = 30^{\circ}$ , $P Q = 2 a$ aur $Q S = a$ . Pythagoras theorem se,

\begin{aligned} P S^{2} & = (2 a)^{2} - a^{2} \\ = 3 a^{2} \\ P S & = \sqrt{3} a \end{aligned}

$∠ Q P S = 30^{\circ}$ se opposite side $Q S = a$ hai aur adjacent side $P S = \sqrt{3} a$ . Usi triangle mein $∠ Q = 60^{\circ}$ se dono chhoti sides ke roles badal jaate hain: opposite side $P S$ hai aur adjacent side $Q S$ . Dono ke liye hypotenuse $P Q = 2 a$ hai.

Ratio	Sides	Value
$\sin 30^{\circ}$	$\frac{a}{2 a}$	$\frac{1}{2}$
$\cos 30^{\circ}$	$\frac{\sqrt{3} a}{2 a}$	$\frac{\sqrt{3}}{2}$
$\tan 30^{\circ}$	$\frac{a}{\sqrt{3} a}$	$\frac{1}{\sqrt{3}}$
$\sin 60^{\circ}$	$\frac{\sqrt{3} a}{2 a}$	$\frac{\sqrt{3}}{2}$
$\cos 60^{\circ}$	$\frac{a}{2 a}$	$\frac{1}{2}$
$\tan 60^{\circ}$	$\frac{\sqrt{3} a}{a}$	$\sqrt{3}$

Reciprocal ratios turant nikal aate hain; woh neeche Table 8.4 mein ikatthe diye gaye hain. Jaise, $cosec 30^{\circ}$ 2 hai, kyunki $\sin 30^{\circ}$ aadha hai.

0° aur 90° ke ratios

Right triangle mein sach mein $0^{\circ}$ ya $90^{\circ}$ ka acute angle nahi ho sakta, isliye definitions seedhe use nahi ho saktin. Iske bajaye hum dekhte hain ki jab angle in values ke bahut kareeb pahunchta hai to ratios ka kya hota hai, aur usi hisaab se ratios define karte hain.

Socho, fixed length ki ek ladder deewar se tiki hai (Fig. 8.7). Ladder hypotenuse hai. Deewar par woh jitni oonchai tak pahunchti hai, woh us angle $A$ ki opposite side hai jo ladder farsh ke saath banati hai, aur deewar se uske neeche ke sire ki distance adjacent side hai. To $\sin A = \frac{height}{ladder}$ aur $\cos A = \frac{base}{ladder}$ .

2/4 $30^{\circ}$ par height bilkul ladder ki aadhi hai: $\sin 30^{\circ} = 0.5$ .

Fig. 8.7Jaise-jaise ladder seedhi khadi hoti hai,

\sin A

badhkar 1 ki taraf jaata hai aur

\cos A

ghatkar 0 ki taraf.NCERT Figs. 8.16–8.18 · p. 123

Fixed length ki ek ladder deewar se tiki hai. Deewar par uski oonchai orange mein mark hai aur us par height likha hai, aur farsh par uska base blue mein mark hai aur us par base likha hai. Ladder ke paas ek readout farsh ke saath angle A, sin A aur cos A dikhata hai. Step one: A = 10°, ladder lagbhag flat, sin A = 0.17 aur cos A = 0.98. Step two: A = 30°, sin A = 0.50 aur cos A = 0.87. Step three: A = 60°, sin A = 0.87 aur cos A = 0.50. Step four: A = 80°, ladder lagbhag seedhi khadi, sin A = 0.98 aur cos A = 0.17.

Jaise-jaise neeche ka sira deewar se door khiskta hai, ladder zyada flat hoti jaati hai aur $A$ ghatkar $0^{\circ}$ ki taraf jaata hai. Height ghatkar 0 ki taraf jaati hai, jabki base badhkar ladder ki poori length ki taraf. To $\sin A$ , 0 ke aur kareeb aata jaata hai, aur $\cos A$ , 1 ke. Isiliye mathematicians define karte hain

\begin{aligned} \sin 0^{\circ} & = 0 \\ \cos 0^{\circ} & = 1 \end{aligned}

$0^{\circ}$ ke baaki ratios isse nikalte hain: $\tan 0^{\circ} = \frac{0}{1} = 0$ aur $\sec 0^{\circ} = \frac{1}{1} = 1$ . Lekin $cosec 0^{\circ}$ aur $\cot 0^{\circ}$ ke liye $\sin 0^{\circ} = 0$ se divide karna padega, jo impossible hai. Hum kehte hain ki yeh defined nahi hain.

Ab ladder ke neeche ke sire ko deewar ki taraf dhakelo. Jaise-jaise $A$ badhkar $90^{\circ}$ ki taraf jaata hai, ladder lagbhag seedhi khadi ho jaati hai: height poori ladder ke kareeb pahunchti hai aur base ghatkar 0 ki taraf. To hum define karte hain

\begin{aligned} \sin 90^{\circ} & = 1 \\ \cos 90^{\circ} & = 0 \end{aligned}

Tab $cosec 90^{\circ} = 1$ aur $\cot 90^{\circ} = \frac{0}{1} = 0$ , jabki $\tan 90^{\circ}$ aur $\sec 90^{\circ}$ ke liye $\cos 90^{\circ} = 0$ se divide karna padega, isliye yeh defined nahi hain.

Saari special values

Yahan saari values ek saath hain, turant dekhne ke liye. Reciprocal ratios doosri table mein hain.

Table 8.3Special angles ke sine, cosine aur tangentNCERT Table 8.1 · p. 125

$A$	$\sin A$	$\cos A$	$\tan A$
$0^{\circ}$	$0$	$1$	$0$
$30^{\circ}$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{3}}$
$45^{\circ}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{\sqrt{2}}$	$1$
$60^{\circ}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\sqrt{3}$
$90^{\circ}$	$1$	$0$	defined nahi

Table 8.4Special angles ke cosecant, secant aur cotangentNCERT Table 8.1 · p. 125

$A$	$cosec A$	$\sec A$	$\cot A$
$0^{\circ}$	defined nahi	$1$	defined nahi
$30^{\circ}$	$2$	$\frac{2}{\sqrt{3}}$	$\sqrt{3}$
$45^{\circ}$	$\sqrt{2}$	$\sqrt{2}$	$1$
$60^{\circ}$	$\frac{2}{\sqrt{3}}$	$2$	$\frac{1}{\sqrt{3}}$
$90^{\circ}$	$1$	defined nahi	$0$

Pehli table mein do trends saaf dikhte hain (Fig. 8.8). $0^{\circ}$ se $90^{\circ}$ ki taraf badhte hue sine dheere-dheere 0 se 1 tak chadhta hai, jabki cosine 1 se 0 tak utarta hai; dono sirf $45^{\circ}$ par milte hain. Tangent bhi chadhta hai, aur $90^{\circ}$ ke paas woh tumhare bataye kisi bhi number se aage nikal jaata hai.

3/3Dono bars sirf $45^{\circ}$ par match karti hain, jahan dono $\frac{1}{\sqrt{2}} \approx 0.71$ hain.

Fig. 8.8Jaise-jaise

A

badhta hai,

\sin A

badhta hai aur

\cos A

ghatta hai.NCERT p. 125

Ek bar chart jiske neeche angles 0°, 30°, 45°, 60° aur 90° hain aur side mein 0 se 1 tak values. Step one sin A ke liye solid blue bars dikhata hai, jo 0, 0.5, 0.71, 0.87 aur 1 tak badhti hain. Step two cos A ke liye hatched orange bars jodta hai, jo 1, 0.87, 0.71, 0.5 aur 0 tak ghatti hain, jabki sine bars halki pad jaati hain. Step three dono sets ki bars poori strength mein dikhata hai aur 45° mark karta hai, jahan dono bars ki height same hai.

Special values ka use

Yeh values pata hon, to tum right triangle ki unknown sides aur angles bina naape nikaal sakte ho.

3/3 $\tan 30^{\circ} = \frac{1.8}{d}$ se $d = 1.8 \sqrt{3} \approx 3.12 m$ milta hai.

Fig. 8.9Right triangle ka baaki sab nikaalne ke liye ek side aur ek angle kaafi hain.NCERT Fig. 8.19 · p. 125

Ek playground slide jiska platform zameen se 1.8 m upar hai aur seedhi slide zameen se 30° par milti hai, isliye slide, platform ka post aur zameen ek right triangle banate hain. Pehle step mein slide ki length aur zameen par distance question marks se mark hain. Doosre step mein slide par 3.6 m likha jaata hai, jo sin 30° se mila. Teesre step mein zameen ki distance par lagbhag 3.12 m likha jaata hai, jo tan 30° se mila.

Example 8.6NCERT Example 6 · p. 125

Slide kitni lambi hai?

Playground slide ka platform zameen se 1.8 m upar hai, aur seedhi slide zameen ke saath $30^{\circ}$ ka angle banati hai (Fig. 8.9). Slide ki length aur platform ke post se uske neeche ke sire ki distance nikaalo.

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

Example 8.7NCERT Example 7 · p. 126

Do sides se angles nikaalna

Ek rassi 5 m oonche pole ke top se seedhi level zameen mein gadi ek khoonti tak jaati hai, jo pole ke base se $5 \sqrt{3}$ m door hai. Rassi zameen ke saath jo angle banati hai aur pole ke saath jo angle banati hai, dono nikaalo.

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

Dekho Example 8.6 aur Example 8.7 ko kya chahiye tha: ek length, aur ek aur jaankari, pehle mein ek angle aur doosre mein ek doosri length. Kyunki right angle pehle se pata hai, right triangle ko poori tarah fix karne ke liye itna hamesha kaafi hai; baaki har side aur angle isse nikal aata hai.

Example 8.8NCERT Example 8 · p. 126

Do equations se do angles

Angles $A$ aur $B$ nikaalo, jahan $A > B$ aur $0^{\circ} < A + B \leq 90^{\circ}$ , jinke liye $\cos (A + B) = 0$ aur $\tan (A - B) = \frac{1}{\sqrt{3}}$ .

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

Quick check

4 mein se 1

\tan 60^{\circ}

kya hai?

\frac{1}{\sqrt{3}}

\sqrt{3}

\frac{\sqrt{3}}{2}

\frac{1}{2}

Answer dekho

Answer (B) $\sqrt{3}$

Aadhe equilateral triangle mein

60^{\circ}

ke opposite side

\sqrt{3} a

hai aur adjacent side

a

, isliye

\tan 60^{\circ} = \sqrt{3}

. Value

\frac{1}{\sqrt{3}}

\tan 30^{\circ}

hai: wahi do sides, ulti taraf se.

8.4 Trigonometric Identities

Algebra mein identity aisi equation hai jo apne variables ki har value ke liye sahi ho, jaise $(x + 1)^{2} = x^{2} + 2 x + 1$ . $x + 1 = 5$ jaisi ordinary equation sirf ek value ke liye sahi hoti hai.

Definition

Trigonometric identity

Trigonometric ratios ki aisi equation jo har us angle ke liye sahi ho jahan uski dono sides defined hon.

Example $\tan A = \frac{\sin A}{\cos A}$ har us $A$ ke liye sahi hai jahan $0^{\circ} \leq A < 90^{\circ}$ .

Tum do identities pehle hi dekh chuke ho, (8.1) aur (8.2). Sabse important identity seedhe Pythagoras theorem se aati hai, aur do aur usi se nikalti hain.

Ek theorem se teen identities

$△ A B C$ lo, jo $B$ par right-angled hai. Pythagoras theorem kehta hai

A B^{2} + B C^{2} = A C^{2}

(8.3)

(8.3) ko $A C^{2}$ se divide karo. Ab har fraction angle $A$ ke ek ratio ka square hai:

\frac{A B^{2}}{A C^{2}} + \frac{B C^{2}}{A C^{2}} = 1

\sin^{2} A + \cos^{2} A = 1

(8.4)

$A = 0^{\circ}$ par yeh $0 + 1 = 1$ ban jaata hai, aur $A = 90^{\circ}$ par $1 + 0 = 1$ . To (8.4) saare $A$ ke liye sahi hai jahan $0^{\circ} \leq A \leq 90^{\circ}$ .

(8.3) ko $A B^{2}$ se divide karo:

1 + \frac{B C^{2}}{A B^{2}} = \frac{A C^{2}}{A B^{2}}

1 + \tan^{2} A = \sec^{2} A

(8.5)

Dono ends check karo: $0^{\circ}$ par dono sides 1 ke barabar hain, lekin $90^{\circ}$ par na $\tan A$ exist karta hai na $\sec A$ . To (8.5) $0^{\circ} \leq A < 90^{\circ}$ ke liye valid hai.

(8.3) ko $B C^{2}$ se divide karo:

\frac{A B^{2}}{B C^{2}} + 1 = \frac{A C^{2}}{B C^{2}}

\cot^{2} A + 1 = {cosec}^{2} A

(8.6)

Is baar problem doosre end par hai: $0^{\circ}$ par $\cot A$ aur $cosec A$ exist nahi karte, isliye (8.6) $0^{\circ} < A \leq 90^{\circ}$ ke liye valid hai.

Teenon ko ek saath dekhne ka ek badhiya tarika hai. Triangle ko is tarah scale karo ki uska hypotenuse 1 ho; tab uski chhoti sides $\cos A$ aur $\sin A$ ban jaati hain. Use is tarah scale karo ki adjacent side 1 ho, ya opposite side 1 ho, to sides baaki ratios ban jaati hain (Fig. 8.10). In har triangle mein Pythagoras theorem ek identity hai.

1/3Hypotenuse 1: legs $\cos A$ aur $\sin A$ hain, isliye $\cos^{2} A + \sin^{2} A = 1$ .

Fig. 8.10Ek triangle, teen sizes, teen identities.NCERT Fig. 8.21 · p. 128

Ek right triangle jiske left corner par angle A hai, teen sizes mein bana hai. Step one: hypotenuse 1 hai, isliye base cos A aur height sin A hai, aur Pythagoras theorem cos²A + sin²A = 1 ban jaata hai. Step two: triangle ko itna bada kiya jaata hai ki base 1 ho jaaye, to height tan A aur hypotenuse sec A hai, jisse 1 + tan²A = sec²A milta hai. Step three: use itna bada kiya jaata hai ki height 1 ho jaaye, to base cot A aur hypotenuse cosec A hai, jisse cot²A + 1 = cosec²A milta hai.

Identities ka use

Identities ek ratio se saare ratios nikaalne ka doosra tarika deti hain, bina triangle draw kiye. Maan lo acute angle $A$ ke liye $\cot A = 2 \sqrt{2}$ . (8.6) se,

\begin{aligned} {cosec}^{2} A & = 1 + \cot^{2} A \\ = 1 + 8 = 9 \end{aligned}

To $cosec A = 3$ (positive, kyunki $A$ acute hai) aur $\sin A = \frac{1}{3}$ . (8.4) se $\cos A = \sqrt{1 - \frac{1}{9}} = \frac{2 \sqrt{2}}{3}$ . Aakhir mein $\tan A = \frac{1}{2 \sqrt{2}}$ aur $\sec A = \frac{3}{2 \sqrt{2}}$ .

Example 8.9NCERT Example 9 · p. 129

Sab kuch $\cos A$ ke terms mein

$0^{\circ}$ aur $90^{\circ}$ ke beech ke angle $A$ ke liye $\sin A$ , $\tan A$ aur $cosec A$ ko sirf $\cos A$ use karke likho.

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

Example 8.10NCERT Example 10 · p. 129

Difference of squares ka use

Prove karo ki $(1 + \cot^{2} A) (1 - \cos A) (1 + \cos A) = 1$ .

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

Example 8.11NCERT Example 11 · p. 130

Common factor bahar nikaalo

Prove karo ki $\frac{\cos A + \cos A \tan^{2} A}{\sin A + \sin A \cot^{2} A} = \tan A$ .

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

Example 8.12NCERT Example 12 · p. 130

Identity se kaam karwao

(a) $\sec^{2} θ - \tan^{2} θ = 1$ use karke dikhao ki $\sec θ - \tan θ = \frac{1}{\sec θ + \tan θ}$ . (b) Agar $\sec θ + \tan θ = 5$ , to $\sec θ$ , $\tan θ$ aur $\sin θ$ nikaalo.

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

Quick check

4 mein se 1

In mein se kaunsi identity nahi hai (un angles ke liye jahan dono sides defined hain)?

\sin^{2} A + \cos^{2} A = 1

\sec^{2} A - \tan^{2} A = 1

{cosec}^{2} A - \cot^{2} A = 1

\sin A + \cos A = 1

Answer dekho

Answer (D) $\sin A + \cos A = 1$

45^{\circ}

par

\sin A + \cos A = \frac{2}{\sqrt{2}} = \sqrt{2}

, 1 nahi.

\sin A

aur

\cos A

ke squares ka sum 1 hota hai.

8.5 Summary

Recap

Summary aur key points

Sides ke naam us angle se do jahan se tum dekh rahe ho: opposite, adjacent, aur right angle ke saamne hypotenuse (Section 8.2).
$\sin A = \frac{opp}{hyp}$ , $\cos A = \frac{adj}{hyp}$ , $\tan A = \frac{opp}{adj}$ ; cosec, sec aur cot inke reciprocals hain (Table 8.1).
$\tan A = \frac{\sin A}{\cos A}$ aur $\cot A = \frac{\cos A}{\sin A}$ ((8.1)).
Ratio sirf angle par depend karta hai, triangle ke size par nahi (similar triangles).
Acute angle ke ek ratio se Pythagoras theorem baaki paanch de deta hai (ek ratio se saare chhah).
Acute $A$ ke liye $\sin A$ aur $\cos A$ , 0 aur 1 ke beech hote hain, $\sec A$ aur $cosec A$ , 1 se zyada hote hain, aur $\tan A$ koi bhi positive number ho sakta hai (ranges).
Aadha square $45^{\circ}$ ke ratios deta hai; aadha equilateral triangle $30^{\circ}$ aur $60^{\circ}$ ke (Table 8.3).
$\sin 0^{\circ} = 0$ , $\cos 0^{\circ} = 1$ , $\sin 90^{\circ} = 1$ , $\cos 90^{\circ} = 0$ ; $\tan 90^{\circ}$ , $\sec 90^{\circ}$ , $cosec 0^{\circ}$ aur $\cot 0^{\circ}$ defined nahi hain (0° aur 90°).
$0^{\circ}$ se $90^{\circ}$ tak $\sin A$ , 0 se badhkar 1 hota hai aur $\cos A$ , 1 se ghatkar 0 (Fig. 8.8).
Ek side aur ek aur part right triangle ko fix kar dete hain (Example 8.6).
$\sin^{2} A + \cos^{2} A = 1$ ; $0^{\circ} \leq A < 90^{\circ}$ ke liye $1 + \tan^{2} A = \sec^{2} A$ ; $0^{\circ} < A \leq 90^{\circ}$ ke liye $\cot^{2} A + 1 = {cosec}^{2} A$ (identities).

1/3Opposite, adjacent aur hypotenuse se $\sin A$ , $\cos A$ aur $\tan A$ milte hain.

Fig. 8.11Poora chapter ek card par.

Teen steps ka revision card. Step one ek right triangle dikhata hai jismein angle A hai, uski sides par adj (blue), opp (orange) aur hyp (purple) likha hai, aur paas mein sin A = opp/hyp, cos A = adj/hyp aur tan A = opp/adj. Step two woh do triangles dikhata hai jo special values dete hain: aadha square jiske legs 1 aur 1, hypotenuse root 2 aur angles 45° hain, aur aadha equilateral triangle jiski sides 1, root 3 aur 2 hain aur angles 30° aur 60°. Step three triangle ko halka karta hai aur identities sin²A + cos²A = 1, 1 + tan²A = sec²A aur 1 + cot²A = cosec²A ki list deta hai.

Flash cards · 10

Khud ko test karo

1 / 10

Quick revision

3 mein se 1

A = 90^{\circ}

par in mein se kaunsa defined nahi hai?

\sin A

\sec A

\cot A

cosec A

Answer dekho

Answer (B) $\sec A$

\sec 90^{\circ}

hota

\frac{1}{\cos 90^{\circ}} = \frac{1}{0}

. Baaki sab theek hain:

\sin 90^{\circ} = 1

\cot 90^{\circ} = 0

aur

cosec 90^{\circ} = 1

Board-exam practice

Yeh questions CBSE board paper ke formats follow karte hain: multiple choice, assertion–reason, short answers, ek proof 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 8.1, Q1	Practice 8.1 · Q1
NCERT Exercise 8.1, Q2	Practice 8.1 · Q2
NCERT Exercise 8.1, Q3	Practice 8.1 · Q3
NCERT Exercise 8.1, Q4	Practice 8.1 · Q4
NCERT Exercise 8.1, Q5	Practice 8.1 · Q5
NCERT Exercise 8.1, Q6	Practice 8.1 · Q6
NCERT Exercise 8.1, Q7	Practice 8.1 · Q7
NCERT Exercise 8.1, Q8	Practice 8.1 · Q8
NCERT Exercise 8.1, Q9	Practice 8.1 · Q9
NCERT Exercise 8.1, Q10	Practice 8.1 · Q10
NCERT Exercise 8.1, Q11	Practice 8.1 · Q11
NCERT Exercise 8.1, Q11 (iii)	Practice 8.1 · Q12
NCERT Exercise 8.2, Q1 (i)–(iv)	Practice 8.2 · Q1
NCERT Exercise 8.2, Q1 (v)	Practice 8.2 · Q2
NCERT Exercise 8.2, Q2	Practice 8.2 · Q3
NCERT Exercise 8.2, Q3	Practice 8.2 · Q4
NCERT Exercise 8.2, Q4	Practice 8.2 · Q5
NCERT Exercise 8.3, Q1	Practice 8.3 · Q1
NCERT Exercise 8.3, Q2	Practice 8.3 · Q2
NCERT Exercise 8.3, Q3	Practice 8.3 · Q3
NCERT Exercise 8.3, Q4 (i)–(iv)	Practice 8.3 · Q4
NCERT Exercise 8.3, Q4 (vi)–(viii)	Practice 8.3 · Q5
NCERT Exercise 8.3, Q4 (ix)–(x)	Practice 8.3 · Q6
NCERT Exercise 8.3, Q4 (v)	Practice 8.3 · Q7

Quick check: MCQs

8.1 Introduction2 questions
Ratios angle par depend karte hain, size par nahi4 questions
Ek ratio se saare chhah tak3 questions
Special values ka use4 questions
Identities ka use4 questions
Quick revision3 questions

Glossary · 10

Key terms (English + Hindi)

adjacent sideसंलग्न भुजाsanlagn bhuja: Right triangle ke acute angle ke liye hypotenuse ke alaawa woh side jo angle banati hai.Chapter mein pehli baar yahan aaya →

cosecant, secant, cotangent: Sine, cosine aur tangent ke reciprocals: $cosec A = \frac{1}{\sin A}$ , $\sec A = \frac{1}{\cos A}$ , $\cot A = \frac{1}{\tan A}$ .Chapter mein pehli baar yahan aaya →
cosineकोज्याkojya: $\cos A$ = $A$ ki adjacent side ÷ hypotenuse.Chapter mein pehli baar yahan aaya →

hypotenuseकर्णkarn: Right triangle mein right angle ke saamne ki side; yeh sabse lambi side hoti hai.Chapter mein pehli baar yahan aaya →

opposite sideसम्मुख भुजाsammukh bhuja: Right triangle ke acute angle ke liye woh side jo angle ko touch nahi karti.Chapter mein pehli baar yahan aaya →

sineज्याjya: $\sin A$ = $A$ ki opposite side ÷ hypotenuse.Chapter mein pehli baar yahan aaya →

tangent (ratio)स्पर्शज्याsparshjya: $\tan A$ = $A$ ki opposite side ÷ $A$ ki adjacent side; yeh $\frac{\sin A}{\cos A}$ bhi hai.Chapter mein pehli baar yahan aaya →
trigonometric identityत्रिकोणमितीय सर्वसमिकाtrikonmitiya sarvasamika: Trigonometric ratios ki aisi equation jo har us angle ke liye sahi ho jiske liye dono sides defined hain.Chapter mein pehli baar yahan aaya →
trigonometric ratioत्रिकोणमितीय अनुपातtrikonmitiya anupaat: Right triangle ke kisi acute angle ke liye li gayi do sides ka ratio, jaise $\sin A$ ya $\cos A$ .Chapter mein pehli baar yahan aaya →
trigonometryत्रिकोणमितिtrikonmiti: Mathematics ki woh branch jo study karti hai ki triangles ki sides aur angles aapas mein kaise jude hain.Chapter mein pehli baar yahan aaya →

Introduction to Trigonometry: Class 10 Maths Chapter 8 Hinglish mein

Is chapter mein aap kya seekhenge

8.1 Introduction

8.2 Trigonometric Ratios

Chhah ratios

Ratios angle par depend karte hain, size par nahi

Ek ratio se saare chhah tak

Practice 8.1: answers ke saath

8.3 Kuch Specific Angles ke Trigonometric Ratios

45° ke ratios

30° aur 60° ke ratios

0° aur 90° ke ratios

Saari special values

Special values ka use

Practice 8.2: answers ke saath

8.4 Trigonometric Identities

Ek theorem se teen identities

Identities ka use

Practice 8.3: answers ke saath

8.5 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