Hello friend! Aaj hum NCERT class 10th ke Mathematics ke chapter 8 jiska naam hai, Introduction to Trigonometry (Trikonmiti ka Parichay) ke exercise 8.1 ke sabhi question ko solve karna sikhne jaa rahe hai. Iss exercise ke ek - ek question ko hum full details me solve karna sikhenge.
Agar aap iss exercise ke question ko solve karna sikhna chahte hai toh bas iss lesson ko ekdam acche se samjhte huye padhiye. Humara ye lesson aapke liye jarur helpful hoga. Hum puri koshish karenge ki, aapko iss exercise ke kisi bhi question me koi bhi doubt yaa problem naa rhe.
Agar aapko kisi bhi question me kisi bhi jagah kuch bhi doubt aaye toh aap comment karke humse puchh sakte hai. Toh chaliye Aaj ka lesson start karte hai.
NCERT Solutions class 10 Math Exercise 8.1 Introduction to Trigonometry Hindi
Exercise - 8.1
(i) sin A, cos A
(ii) sin C, cos C
Solution-: Diya hai ∆ABC me ∠B = 90°, AB = 24 cm aur BC = 7 cm. (Figure - 1 dekhiye)
 |
| Figure - 1 |
Pythagoras theorem (pramey) ke anusar,
AC² = AB²+BC²
= (24)²+(7)²
= 576+49
= 625
AC = √625
AC = 25
(i) sin A, cos A
Hum jante hai ki, sin A = Lamb/karn
= BC/AC
= 7/25
Issi prakar se, cos A = Aadhar/karn.
= AB/AC
= 24/25
Toh sin A = 7/25 aur cos A = 24/25 Ans.
(ii) sin C, cos C
sin C = Lamb/karn
= AB/AC
= 24/25
Aur cos C = Aadhar/karn
= BC/AC
= 7/25
Toh sin C = 24/25 aur cos C = 7/25 Ans.
(2). Figure - 2 me, tan P - cot R ka maan gyaat kijiye.
 |
| Figure - 2 |
Solution-: Figure - 2 me, diya hai Samkon ∆PQR jisme ∠Q = 90°, PQ = 12 cm aur PR = 13 cm.
PR² = PQ² + QR²
169 = 144 + QR²
QR² = 169 - 144
QR² = 25
QR = √25
QR = 5
Hum jaante hai ki, tan P = Lamb/Aadhar
= QR/PQ
= 5/12
Iss prakar se, cot R = Aadhar/Lamb
= QR/PQ
= 5/12
Toh, tan P - cot R = 5/12 - 5/12
= 0
So, tan P - cot R = 0 Ans.
(3). Yadi sin A = 3/4, toh cos A aur tan A ka maan parikalit kijiye.
Solution-: Diya hai, sin A = 3/4
Hum jante hai ki, sin A = Lamb/karn
∴ Lamb = 3, karn = 4 aur Aadhar = ??
Maana Samkon ∆ABC me, ∠B = 90°, BC (lamb) = 3, AC (Karn) = 4 aur AB (aadhar) = ?? (Figure - 3 me dekhiye)
 |
| Figure - 3 |
Toh Pythagoras pramey ke anusar,
AC² = AB² + BC²
4² = AB² + 3²
16 = AB² + 9
AB² = 16 - 9
AB = √7
Toh cos A = Aadhar/Karn
= AB/AC
= √7/4
Aur tan A = lamb/Aadhar
= BC/AB
= 3/√7
Toh cos A = √7/4 aur tan A = 3/√7 Ans.
(4). Yadi 15cot A = 8 ho toh sin A aur sec A ka maan gyaat kijiye.
Solution-: Diya hai, 15cot A = 8
⇒ cot A = 8/15
Hum jante hai ki, cot A = Aadhar/Lamb
∴ Aadhar = 8, lamb = 15 aur karn = ??
Maana Samkon ∆ABC me, ∠B = 90°, AB (aadhar) = 8, BC (lamb) = 15 aur AC (Karn) = ?? (Figure - 4 me dekhiye)
 |
| Figure - 4 |
Toh Pythagoras pramey ke anusar,
AC² = AB² + BC²
AC² = 8² + 15²
AC² = 64 + 225
AC² = 289
AC = √289
AC = 17
Toh sin A = Lamb/Karn
= BC/AC
= 15/17
Aur sec A = Karn/aadhar
= AC/AB
= 17/8
Toh, sin A = 15/17 aur sec A = 17/8 Ans.
NCERT solutions for class 10th Math ex 8.1 Question- 5 to 8 in Hindi
(5). Yadi sec θ = 13/12, ho toh shesh sabhi tirkonmitiy anupaat parikalit kijiye.
Solution-: Diya hai, sec θ = 13/12
Hum jante hai ki, sec θ = Karn/Aadhar
∴ Karn = 13 aur Aadhar = 12
Maana Samkon ∆ABC me, ∠B = 90°, ∠A = θ, AB (aadhar) = 12, AC (karn) = 13 aur BC (lamb) = ?? (Figure - 5 me dekhiye)
 |
| Figure - 5 |
Toh Pythagoras pramey ke anusar,
AC² = AB² + BC²
13² = 12² + BC²
169 = 144 + BC²
BC² = 169 - 144
BC² = 25
BC = √25
BC = 5
Toh sin θ = Lamb/karn
= BC/AC
= 5/13
cos θ = Aadhaar/karn
= AB/AC
= 12/13
tan θ = Lamb/Aadhar
= BC/AB
= 5/12
cosec θ = Karn/Lamb
= AC/BC
= 13/5
cot θ = Aadhaar/Lamb
= AB/BC
= 12/5
(6). Yadi ∠A aur ∠B nyunkon (acute angle) ho, jaha cos A = cos B, toh dikhaiye ki ∠A = ∠B
Solution-: Maana Samkon ∆ABC me, ∠C = 90° hai, Diya hai, ∠A aur ∠B nyunkon (acute angle) hai. (Figure - 6 me dekhiye)
 |
| Figure - 6 |
∆ABC se, cos A = AC/AB aur cos B = BC/AB
∵ cos A = cos B (Diya hai.)
⇒ AC/AB = BC/AB
⇒ AC = BC
⇒ ∠A = ∠B [ Triangle ke equal sides ke opposite angles equal hote hai.]
(7). Yadi cot θ = 7/8, toh
(ii) cot² θ ka maan nikaaliye?
Solution-: Diya hai, cot θ = 7/8
Hum jante hai ki, cot θ = Aadhar/Lamb
Isliye, Aadhar = 7 aur Lamb = 8
Maana Samkon ∆ABC me, ∠B = 90°, ∠A = θ, AB (aadhar) = 7, BC (lamb) = 8 aur AC (karn) = ? (Figure - 7 me dekhiye)
 |
| Figure - 7 |
Toh ∆ABC me Pythagoras pramey ke anusar,
AC² = AB² + BC²
AC² = 7² + 8²
AC² = 49 + 64
AC² = 113
AC = √113
Toh sin θ = Lamb/karn
= BC/AC
= 8/√113
cos θ = Aadhaar/karn
= AB/AC
= 7/√113
(ii) cot² θ ka maan nikaaliye?
Diya hai ki, cot θ = 7/8
⇒ cot² θ = (7/8)²
= 49/84 Ans.
(8). 3 cot A = 4, toh jaanch kijiye ki
Solution -: Diya hai, 3 cot A = 4
⇒ cot A = 4/3
Hum jante hai ki, cot A = Aadhar/Lamb
∴ Aadhar = 4 aur Lamb = 3
Maana samkon ∆ABC me, ∠B = 90°, AB (aadhar) = 4, BC (lamb) = 3 aur AC (karn) = ? (Figure - 8 me dekhiye)
 |
| Figure - 8 |
Toh Pythagoras pramey ke anusar,
AC² = AB² + BC²
AC² = 4² + 3²
AC² = 16 + 9
AC² = 25
AC = √25
AC = 5
Toh sin A = Lamb/Karn
= BC/AC
= 3/5
cos A = Aadhar/Karn
= AB/AC
= 4/5
aur tan A = Lamb/Aadhar
= BC/AB
= 3/4
L.H.S. = 1 - tan² A
1 + tan² A
= 1 - (3/4)²
1 + (3/4)²
= 1 - 9/16
1 + 9/16
= (16 - 9)/16
(16 + 9)/16
= 7/16
25/16
= 7/16 × 16/25
= 7/25
R.H.S. = cos² A - sin² A
= (4/5)² - (3/5)²
= 16/25 - 9/25
= (16-9)/25
= 7/25
Toh, 1 - tan² A = cos² A - sin² A hai.
1 + tan² A
NCERT solutions for class 10th Math ex 8.1 Question- 9 to 11 in Hindi
(i) sin A cos C + cos A sin C
(ii) cos A cos C - sin A sin C
Solution-: Diya hai, Samkon ∆ABC me, ∠B = 90° aur tan A = 1/√3
Hum jante hai ki, tan A = Lamb/Aadhar
∴ Lamb = 1 aur Aadhar = √3
Maana ∆ABC me Lamb (BC) = 1, Aadhar (AB) = √3 aur Karn (AC) = ?? (Figure - 9 me dekhiye)
 |
| Figure - 9 |
Toh ∆ABC me Pythagoras pramey ke anusar,
AC² = AB² + BC²
AC² = (√3)² + 1²
AC² = 3 + 1
AC² = 4
AC = √4
AC = 2
Toh sin A = Lamb/Karn
= BC/AC
= 1/2
sin C = AB/AC
= √3/2
Issi prakar se, cos A = Aadhar/Karn
= AB/AC
= √3/2
aur cos C = BC/AC
= 1/2
(i) sin A cos C + cos A sin C
= 1/2 × 1/2 + √3/2 × √3/2
= 1/4 + 3/4
= (1+3)/4
= 4/4
= 1 Ans.
(ii) cos A cos C - sin A sin C
= √3/2 × 1/2 - 1/2 × √3/2
= √3/4 - √3/4
= 0 Ans.
(10). ∆PQR me, jiska kon Q samkon hai, PR + QR = 25 cm aur PQ = 5 cm hai. sin P, cos P aur tan P ke maan gyaat kijiye.
Solution-: Diya hai: ∆PQR me ∠Q = 90°, PR+QR = 25cm aur PQ = 5cm . (Figure - 10 me dekhiye)
 |
| Figure - 10 |
Maana QR = x cm
Toh PR+QR = 25cm
⇒ PR = 25 - QR
⇒ PR = 25 - x
∆ABC me Pythagoras pramey se,
PR² = QR² + PQ²
⇒(25 - x)² = QR² + 5²
⇒(25)² + x² - 2×25×x = QR² + 25
⇒625 + x² -50x = x² + 25
⇒625 + x² - x² - 25 = 50x
⇒(625 - 25) + (x² - x²) = 50x
⇒600 = 50x
⇒x = 600/50
⇒x = 12
⇒QR = 12cm
Toh PR+QR = 25cm
PR + 12 = 25
PR = 25 - 12
PR = 13cm
∴ PQ = 5cm, QR = 12cm aur PR = 13cm
sin P = Lamb/Karn
= QR/PR
= 12/13
cos P = Aadhar/Karn
= PQ/PR
= 5/13
tan P = Lamb/Aadhar
= QR/PQ
= 12/5
sin P = 12/13, cos P = 5/13 aur tan P = 12/5 Ans.
(11). Bataaiye ki nimnlikhit kathan Sahi (True) hai ya Galat (False). Kaaran sahit apne answer ki pushti kijiye.
(i) tan A ka maan humesha 1 se kam hota hai.
(ii) Kon A ke kisi maan ke liye sec A = 12/5
(iii) cos A, kon A ke cosecant ke liye short form hai.
(iv) cot A, cot aur A ka gunanphal hota hai.
(v) Kisi bhi kon θ ke liye sin θ = 4/3
Answer-(i). Galat (False)
Proof-: Maana Samkon ∆ABC, jisme ∠B = 90°, AB = 3 aur BC = 4 hai. (Figure me dekhiye)
Toh Pythagoras pramey ke anusar,
AC² = AB² + BC²
AC² = 3² + 4²
AC² = 9 + 16
AC² = 25
AC = √25
AC = 5
∆ABC se, tan A = Lamb/Aadhar
= BC/AB
= 4/3
= 1.3 (approx.)
Hum dekh sakte hai ki, 1.3 > 1
Phir ∆ABC se, tan C = Lamb/Aadhar
= AB/BC
= 3/4
= 0.75
Yaha hum dekh sakte hai ki, 0.75<1
∴ tan A ka maan 1 se Chhota (small) aur bada (greater) aur equal ho sakta hai.
Answer-(ii). Sahi (True)
Proof-: Diya hai sec A = 12/5
Hum jante hai ki, sec A = Karn/Aadhaar
Toh karn = 12 aur Aadhar = 5
Maana Samkon ∆ABC, jisme ∠B = 90°, AB (Aadhar) = 5 aur AC (karn) = 12 (Figure me dekhiye)
Toh Pythagoras pramey ke anusar,
AC² = AB² + BC²
12² = 5² + BC²
144 = 25 + BC²
BC² = 144 - 25
BC² = 119
BC = √119
Lamb = √119
Hum dekh sakte hai ki, iss prakar ka Samkon tribhuj (triangle) ho sakta hai, jisme AB (Aadhar) = 5, BC (Lamb) = √119 aur AC (karn) = 12 isliye sec A = 12/5 sahi hai.
Answer-(iii). Galat (False), kyuki cos A ka short form cosine hota hai.
Answer-(iv). Galat (False), kyuki cot A , pura ek term (single term) hai. Yaa Hum ye bhi kah sakte hai ki, cot A, ∠A ka ek trikonmity anupaat (trigonometric ratio) hai
Answer-(v). Galat (False)
Proof-: Diya hai, sin θ = 4/3
Hum jante hai ki, sin θ = Lamb/karn
Toh Lamb = 4 aur Karn = 3
Hum dekh sakte hai ki, Karn < Lamb
Lekin Hum jante hai ki, Samkon tribhuj me Karn sabse badi (longest) side hoti hai. Isliye sin θ = 4/3 galat hai.
❒ Relation among sinθ, cosθ, tanθ aur cotθ
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