RD Sharma Solutions Class 10 Maths Chapter 5 Exercise 5.3 Trigonometric Ratios
Neha Tanna14 Nov, 2024
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RD Sharma Solutions Class 10 Maths Chapter 5 Exercise 5.3: Chapter 5, Exercise 5.3 in RD Sharma’s Class 10 Maths book focuses on trigonometric ratios, a key concept in trigonometry. This exercise introduces the fundamental ratios—sine, cosine, tangent, cotangent, secant, and cosecant—defined based on the sides of a right triangle.
It guides students to calculate these ratios using given angles and triangle side lengths, helping them understand relationships among these ratios. Through step-by-step problem-solving, students enhance their ability to find and apply trigonometric ratios in various mathematical and real-world contexts.RD Sharma Solutions Class 10 Maths Chapter 5 Exercise 5.3 Overview
Chapter 5, Exercise 5.3 in RD Sharma’s Class 10 Maths emphasizes the importance of trigonometric ratios—sine, cosine, tangent, and their reciprocals (cosecant, secant, and cotangent)—which are fundamental in solving geometry and physics problems. Understanding these ratios allows students to relate the angles of a right triangle to its sides, which is crucial for calculating heights, distances, and slopes in real-world applications, such as navigation, architecture, and engineering. Mastery of trigonometric ratios forms the basis for advanced studies in calculus and analytical geometry, providing essential tools for problem-solving in various scientific and mathematical fields.RD Sharma Solutions Class 10 Maths Chapter 5 Exercise 5.3 PDF
Below, we have provided the PDF for RD Sharma Solutions for Class 10 Maths, Chapter 5, Exercise 5.3: Trigonometric Ratios. This resource offers clear, step-by-step solutions to help students understand trigonometric concepts and problem-solving techniques. These solutions are beneficial for building a solid foundation in trigonometry, essential for board exams and further studies in mathematics.RD Sharma Solutions Class 10 Maths Chapter 5 Exercise 5.3 PDF
RD Sharma Solutions Class 10 Maths Chapter 5 Exercise 5.3 Trigonometric Ratios
Below is the RD Sharma Solutions Class 10 Maths Chapter 5 Exercise 5.3 Trigonometric Ratios -1. Evaluate the following:
(i) sin 20 o / cos 70 o
(ii) cos 19 o / sin 71 o
(iii) sin 21 o / cos 69 o
(iv) tan 10 o / cot 80 o
(v) sec 11 o / cosec 79 o
Solution:
(i) We have, sin 20 o / cos 70 o = sin (90 o – 70 o )/ cos 70 o = cos 70 o / cos70 o = 1 [∵ sin (90 – θ) = cos θ] (ii) We have, cos 19 o / sin 71 o = cos (90 o – 71 o )/ sin 71 o = sin 71 o / sin 71 o = 1 [∵ cos (90 – θ) = sin θ] (iii) We have, sin 21 o / cos 69 o = sin (90 o – 69 o )/ cos 69 o = cos 69 o / cos69 o = 1 [∵ sin (90 – θ) = cos θ] (iv) We have, tan 10 o / cot 80 o = tan (90 o – 10 o ) / cot 80 o = cot 80 o / cos80 o = 1 [∵ tan (90 – θ) = cot θ] (v) We have, sec 11 o / cosec 79 o = sec (90 o – 79 o )/ cosec 79 o = cosec 79 o / cosec 79 o = 1 [∵ sec (90 – θ) = cosec θ]2. Evaluate the following:
Solution:
We have, [∵ sin (90 – θ) = cos θ and cos (90 – θ) = sin θ]
= 1
2
+ 1
2
= 1 + 1
= 2
(ii) cos 48°- sin 42°
Solution:
We know that cos (90° − θ) = sin θ. So, cos 48° – sin 42° = cos (90° − 42°) – sin 42° = sin 42° – sin 42°= 0 Thus, the value of cos 48° – sin 42° is 0.
Solution:
We have, [∵ cot (90 – θ) = tan θ and cos (90 – θ) = sin θ]
= 1 – 1/2(1)
= 1/2
Solution:
We have, [∵ sin (90 – θ) = cos θ and cos (90 – θ) = sin θ]
= 1 – 1
= 0
Solution:
We have, [∵ cot (90 – θ) = tan θ and tan (90 – θ) = cot θ]
= tan (90
o
– 35
o
)/ cot 55
o
+ cot (90
o
– 12
o
)/ tan 12
o
– 1
= cot 55
o
/ cot 55
o
+ tan 12
o
/ tan 12
o
– 1
= 1 + 1 – 1
= 1
Solution:
We have , [∵ sin (90 – θ) = cos θ and sec (90 – θ) = cosec θ]
(vii) cosec 31° – sec 59°
Solution:
We have, cosec 31° – sec 59° Since, cosec (90 – θ) = cos θ So, cosec 31° – sec 59° = cosec (90° – 59 o ) – sec 59° = sec 59° – sec 59° = 0 Thus, cosec 31° – sec 59° = 0(viii) (sin 72° + cos 18°) (sin 72° – cos 18°)
Solution:
We know that, sin (90 – θ) = cos θ So, the given can be expressed as (sin 72° + cos 18°) (sin (90 – 18)° – cos 18°) = (sin 72° + cos 18°) (cos 18° – cos 18°) = (sin 72° + cos 18°) x 0 = 0(ix) sin 35° sin 55° – cos 35° cos 55°
Solution:
We know that, sin (90 – θ) = cos θ So, the given can be expressed as sin (90 – 55)° sin (90 – 35)° – cos 35° cos 55° = cos 55° cos 35° – cos 35° cos 55° = 0(x) tan 48° tan 23° tan 42° tan 67°
Solution:
We know that, tan (90 – θ) = cot θ So, the given can be expressed as tan (90 – 42)° tan (90 – 67)° tan 42° tan 67° = cot 42° cot 67° tan 42° tan 67° = (cot 42° tan 42°)(cot 67° tan 67°) = 1 x 1 [∵ tan θ x cot θ = 1] = 1(xi) sec 50° sin 40° + cos 40° cosec 50°
Solution:
We know that, sin (90 – θ) = cos θ and cos (90 – θ) = sin θ So, the given can be expressed as sec 50° sin (90 – 50)° + cos (90 – 50)° cosec 50° = sec 50° cos 50° + sin 50° cosec 50° = 1 + 1 [∵ sin θ x cosec θ = 1 and cos θ x sec θ = 1] = 23. Express each one of the following in terms of trigonometric ratios of angles lying between 0 o and 45 o
(i) sin 59 o + cos 56 o (ii) tan 65 o + cot 49 o (iii) sec 76 o + cosec 52 o
(iv) cos 78 o + sec 78 o (v) cosec 54 + sin 72 o (vi) cot 85 o + cos 75 o
(vii) sin 67 o + cos 75 o
Solution:
Using the below trigonometric ratios of complementary angles, we find the required sin (90 – θ) = cos θ cosec (90 – θ) = sec θ cos (90 – θ) = sin θ sec (90 – θ) = cosec θ tan (90 – θ) = cot θ cot (90 – θ) = tan θ (i) sin 59 o + cos 56 o = sin (90 – 31) o + cos (90 – 34) o = cos 31 o + sin 34 (ii) tan 65 o + cot 49 o = tan (90 – 25) o + cot (90 -41) o = cot 25 o + tan 41 o (iii) sec 76 o + cosec 52 o = sec (90 – 14) o + cosec (90 – 38) o = cosec 14 + sec 38 o (iv) cos 78 o + sec 78 o = cos (90 – 12) o + sec (90 – 12) o = sin 12 o + cosec 12 o (v) cosec 54 + sin 72 o = cosec (90 – 36) o + sin (90 – 18) o = sec 36 o + cos 18 o (vi) cot 85 o + cos 75 o = cot (90 – 5) o + cos (90 – 15) o = tan 5 o + sin 15 o4. Express cos 75 o + cot 75 o in terms of angles between 0 o and 30 o .
Solution:
Given, cos 75 o + cot 75 o Since, cos (90 – θ) = sin θ and cot (90 – θ) = tan θ cos 75 o + cot 75 o = cos (90 – 15) o + cot (90 – 15) o = sin 15 o + tan 15 o Hence, cos 75 o + cot 75 o can be expressed as sin 15 o + tan 15 o5. If sin 3A = cos (A – 26 o ), where 3A is an acute angle, find the value of A.
Solution:
Given, sin 3A = cos (A – 26 o ) Using cos (90 – θ) = sin θ, we have sin 3A = sin (90 o – (A – 26 o )) Now, comparing both L.H.S and R.H.S 3A = 90 o – (A – 26 o ) 3A + (A – 26 o ) = 90 o 4A – 26 o = 90 o 4A = 116 o A = 116 o /4 ∴ A = 29 o6. If A, B, C are the interior angles of a triangle ABC, prove that
(i) tan ((C + A)/ 2) = cot (B/2) (ii) sin ((B + C)/ 2) = cos (A/2)
Solution:
We know that in triangle ABC, the sum of the angles, i.e., A + B + C = 180 o So, C + A = 180 o – B ⇒ (C + A)/2 = 90 o – B/2 …… (i) And, B + C = 180 o – A ⇒ (B + C)/2 = 90 o – A/2 ……. (ii) (i) L.H.S = tan ((C + A)/ 2) ⇒ tan ((C + A)/ 2) = tan (90 o – B/2) [From (i)] = cot (B/2) [∵ tan (90 – θ) = cot θ] = R.H.S- Hence Proved
- Hence Proved
7. Prove that:
(i) tan 20° tan 35° tan 45° tan 55° tan 70° = 1
(ii) sin 48° sec 48° + cos 48° cosec 42° = 2
Solution:
(i) Taking L.H.S = tan 20° tan 35° tan 45° tan 55° tan 70° = tan (90° − 70°) tan (90° − 55°) tan 45°tan 55° tan70° = cot 70°cot 55° tan 45° tan 55° tan 70° [∵ tan (90 – θ) = cot θ] = (tan 70°cot 70°)(tan 55°cot 55°) tan 45° [∵ tan θ x cot θ = 1] = 1 × 1 × 1 = 1- Hence proved
- Hence proved
= 1 + 1 – 2
= 2 – 2
= 0
- Hence proved
= 1 + 1
= 2
- Hence proved
8. Prove the following:
(i) sinθ sin (90 o – θ) – cos θ cos (90 o – θ) = 0
Solution:
Taking the L.H.S, sinθ sin (90 o – θ) – cos θ cos (90 o – θ) = sin θ cos θ – cos θ sin θ [∵ sin (90 – θ) = cos θ and cos (90 – θ) = sin θ] = 0
(ii)
Solution:
Taking the L.H.S,
[∵ cosec θ x sin θ = 1 and cos θ x sec θ = 1]
= 1 + 1
= 2 = R.H.S
- Hence Proved
(iii)
Solution:
Taking the L.H.S, [∵ tan (90 o – θ) = cot θ]
= 0 = R.H.S
- Hence Proved
(iv)
Solution:
Taking L.H.S, [∵ sin (90 – θ) = cos θ and cos (90 – θ) = sin θ]
= sin
2
A = R.H.S
- Hence Proved
(v) sin (50 o + θ ) – cos (40 o – θ ) + tan 1 o tan 10 o tan 20 o tan 70 o tan 80 o tan 89 o = 1
Solution:
Taking the L.H.S, = sin (50 o + θ) – cos (40 o – θ) + tan 1 o tan 10 o tan 20 o tan 70 o tan 80 o tan 89 o = [sin (90 o – (40 o – θ))] – cos (40 o – θ) + tan (90 – 89) o tan (90 – 80) o tan (90 – 70) o tan 70 o tan 80 o tan 89 o [∵ sin (90 – θ) = cos θ] = cos (40 o – θ) – cos (40 o – θ) + cot 89 o cot 80 o cot 70 o tan 70 o tan 80 o tan 89 o [∵ tan (90 o – θ) = cot θ] = 0 + (cot 89 o x tan 89 o ) (cot 80 o x tan 80 o ) (cot 70 o x tan 70 o ) = 0 + 1 x 1 x 1 [∵ tan θ x cot θ = 1] = 1= R.H.S- Hence Proved
Benefits of Solving RD Sharma Solutions Class 10 Maths Chapter 5 Exercise 5.3
The RD Sharma Solutions for Class 10 Maths, Chapter 5, Exercise 5.3 on Trigonometric Ratios offer several benefits:Concept Clarity : The solutions break down trigonometric concepts step-by-step, helping students grasp fundamental ideas effectively.
Practice and Accuracy : Solving diverse problems enhances accuracy and builds confidence in handling trigonometric calculations.
Exam Preparation : Detailed solutions align with board exam standards, preparing students for various question patterns.
Foundation for Advanced Maths : Mastery in trigonometry is essential for higher-level math studies, making these solutions valuable for long-term learning.
Error-Free Learning : With precise answers, students can check their work and avoid common mistakes.
RD Sharma Solutions Class 10 Maths Chapter 5 Exercise 5.3 FAQs
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