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8. त्रिकोणमिति का परिचय is one of the most important chapters in the Class 10 Mathematics Hindi NCERT Solutions curriculum. This chapter plays a significant role in helping students build a strong conceptual foundation while preparing for school examinations, class tests, unit tests, half-yearly examinations, annual examinations, and CBSE board assessments. The chapter has been carefully designed according to the latest NCERT syllabus, making it an essential part of every student's study plan.
The 8. त्रिकोणमिति का परिचय - Class 10 Mathematics Hindi NCERT Solutions available on ATP Education explain every question in a simple, accurate, and step-by-step manner. Each answer is prepared according to the latest CBSE guidelines so that students can understand the concepts clearly without confusion. Whether you are completing your homework, revising before examinations, or strengthening your understanding of the subject, these solutions provide reliable academic support throughout your learning journey.
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Our Class 10 Mathematics NCERT Solutions cover all textbook questions, important exercise questions, and chapter-wise explanations in Hindi Medium. Every solution is written in easy-to-understand language, allowing students to revise the chapter quickly before examinations. Regular practice of these solutions improves confidence, strengthens subject knowledge, and reduces examination stress.
Students preparing for school assessments should carefully study 8. त्रिकोणमिति का परिचय because questions from this chapter are frequently asked in objective questions, short answer questions, long answer questions, competency-based questions, and case-study questions. Understanding the concepts explained in this chapter also helps students connect related topics from other chapters, making overall learning more effective and meaningful.
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8. त्रिकोणमिति का परिचय - Class 10 Mathematics Hindi NCERT Solutions
8. त्रिकोणमिति का परिचय
प्रश्नावली 8.1 (गणित)
Q1. ΔABC में, जिसका कोण B समकोण है, AB = 24 cm और BC = 7 cm है | निम्न लिखित का मान ज्ञात कीजिए :
(i) sin A, cos A
(ii) sin C, cos C
Solution:
समकोण त्रिभुज ΔABC में,
AB = 24 cm, BC = 7 cm
पाइथागोरस प्रमेय से,
AC2 = AB2 + BC2
= 242 + 72
= 576 + 49
= 625
AC = √625 = 25 cm
अब तत्रिकोणमितिय अनुपात लेने पर
(i) sin A, cos A
Q2. आकृति 8.13 में, tan P – cot R का मान ज्ञात कीजिए |
Solution:
PQ = 12 cm, PR = 13 cm
QR = ?
समकोण त्रिभुज ΔPQR में,
PQ = 12 cm, PR = 13 cm
पाइथागोरस प्रमेय से,
PR2 = PQ2 + QR2
132 = 122 + QR2
169 = 144 + QR2
169 - 144 = QR2
QR2 = 25
QR = √25 = 5 cm
अब तत्रिकोणमितिय अनुपात लेने पर
8. त्रिकोणमिति का परिचय
प्रश्नावली 8.2
Q1. निम्नलिखित के मान निकालिए:
(i) sin 60° cos 30° + sin 30° cos 60°
हल: sin 60° cos 30° + sin 30° cos 60°
सभी त्रिकोंणमितीय अनुपातों का मान रखने पर
(ii) 2 tan2 45° + cos2 30° – sin2 60°
हल: 2 tan2 45° + cos2 30° – sin2 60°
Q2. सही विकल्प चुनिए और अपने विकल्प का औचित्य दीजिए :
(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°
(iii) sin 2A = 2 sin A तब सत्य होता है, जबकि A बराबर है :
(A) 0° (B) 30° (C) 45° (D) 60°
हल: sin 2A = 2 sin A
⇒ 2 sin A cos A = 2 sin A [ sin 2x = 2 sin x cos x]
⇒ cos A = 2 sin A - 2 sin A
⇒ cos A = 0
∴ A = 0o
विकल्प (A) सत्य है |
A का मान समीकरण (iii) में रखने पर
A + B = 60°
⇒ 45° + B = 60°
⇒ B = 60° - 45°
⇒ B = 15°
A = 45°, B = 15°
Q4. बताइए कि निम्नलिखित में से कौन-कौन सत्य हैं या असत्य हैं | कारण सहित अपने उत्तर की पुष्टि कीजिए |
(i) sin (A + B) = sin A + sin B.
(ii) θ में वृद्धि होने के साथ sin θ के मान में भी वृद्धि होती है |
(iii) θ में वृद्धि होने के साथ cos θ के मान में भी वृद्धि होती है |
(iv) θ के सभी मानों पर sin θ = cos θ
(v) A = 0° पर cot A परिभाषित नहीं है |
उत्तर:
(i) दिया गया कथन असत्य है |
(ii) दिया गया कथन सत्य है |
(iii) दिया गया कथन असत्य है |
(iv) दिया गया कथन असत्य है |
(v) दिया गया कथन सत्य है |
8. त्रिकोणमिति का परिचय
अध्याय 8. त्रिकोणमितिय अनुपातों का परिचय
प्रश्नावली 8.3
Q1. निम्नलिखित का मान निकालिए:
(iii) cos 48° - sin 42°
हल: cos 48° - sin 42°
⇒ sin(90° - 48°) - sin 42°
⇒ sin 42° - sin 42° = 0
(iv) cosec 31° - sec 59°
हल: cosec 31° - sec 59°
⇒ sec (90° - 31°) - sec 59° [ cosec q = sec (90° - q) ]
⇒ sec 59° - sec 59° = 0
Q2. दिखाइए कि
(i) tan 48° tan 23° tan 42° tan 67° = 1
हल: (i) tan 48° tan 23° tan 42° tan 67° = 1
LHS = tan 48° tan 23° tan 42° tan 67°
= cot (90° - 48°) tan (90° - 23°) tan 42° tan 67°
= cot 42° cot 67° tan 42° tan 67°
= (cot 42° × tan 42°) (cot 67° × tan 67°)
= 1 × 1 [ cot A × tan A = 1 ]
= 1
LHS = RHS
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
हल: (ii) cos 38° cos 52° – sin 38° sin 52° = 0
LHS = cos 38° cos 52° sin 38° sin 52°
= sin (90° - 38°) cos 52° – cos (90° - 38°) sin 52°
= sin 52° cos 52° - cos 52° sin 52°
= sin 52° (cos 52° - cos 52°)
= sin 52° × 0
= 0
LHS = RHS
Q3. यदि tan 2A = cot(A - 18°), जहाँ 2A एक न्यूनकोण है, तो A का मान ज्ञात कीजिए |
हल: tan 2A = cot(A - 18°),
⇒ cot (90° - 2A) = cot(A - 18°)
दोनों पक्षों में तुलना करने पर
⇒ 90° - 2A = A - 18°
⇒ 90° + 18° = A + 2A
⇒ 3A = 108°
Q4. यदि tan A = cot B, तो सिद्ध कीजिए कि A + B = 90°
हल: tan A = cot B दिया है |
⇒ tan A = tan (90° - B) तुलना करने पर
⇒ A = 90° - B
⇒ A + B = 90° Proved
Q5. यदि sec 4A = cosec(A - 20°), जहाँ 4A एक न्यूनकोण है, तो A का मान ज्ञात कीजिए |
हल: sec 4A = cosec(A - 20°)
⇒ cosec (90° - 4A) = cosec(A - 20°) [ sec q = (90°- q) ]
तुलना करने पर
⇒ 90° - 4A = A - 20°
⇒ 90° + 20° = A + 4A
⇒ 5A = 110°
Q7. sin 67° + cos 75° को 0° और 45° के बीच के कोणों के त्रिकोणमितिय अनुपातों के पदों में व्यक्त कीजिए |
हल : sin 67° + cos 75°
⇒ cos (90° - 67°) + sin (90° - 75°)
⇒ cos 23° + sin 15°
8. त्रिकोणमिति का परिचय
अध्याय 8. त्रिकोणमिति का परिचय
NCERT SOLUTION
अभ्यास 8.4
Q4. सही विकल्प चुनिए और अपने विकल्प की पुष्टि कीजिए:
(i) 9 sec2 A – 9 tan2 A बराबर है:
(A) 1 (B) 9 (C) 8 (D) 0
Correct Answer: (B) 9
Solution:
9 sec2 A – 9 tan2 A = 9(sec2 A – tan2 A)
= 9 × 1 = 9
(ii) (sec A + tan A) (1 – sin A) बराबर हैं :
(A) sec A (B) sin A (C) cosec A (D) cosA
Correct Answer: (D) cosA
Q5. निम्नलिखित सर्वसमिका सिद्ध कीजिए, जहाँ वे कोण, जिनके लिए व्यंजक परिभाषित है, न्यूनकोण है :
अत: LHS = RHS proved
**********The End *********
8. त्रिकोणमिति का परिचय
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