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Selina Solutions Concise Maths Class 10 Chapter 21 Trigonometrical Identities Exercise 21(D)


Selina Solutions Class 10 Maths Chapter 21 Trigonometrical Identities

The science related to the measurement of triangles is called trigonometry. Trigonometric ratios and its relations are used to prove trigonometric identities. In addition, trigonometric ratios of complementary angles and the use of trigonometric tables are other topics covered in this chapter. Since this chapter lays the foundation for high grade mathematics, students should gain a strong grip on this chapter. For this purpose, www.mathspdfsolution.co  has created Selina Solutions for Class 10 Mathematics prepared by expert faculty with vast academic experience. It also improves students' problem-solving skills, which are important from the exam point of view. Selina Solutions for Class 10 Mathematics Chapter 21 Trigonometric Identification PDFs are available practice-wise in the link below.

Exercise 21(A) Solutions

Exercise 21(B) Solutions

Exercise 21(C) Solutions

Exercise 21(D) Solutions

Exercise 21(E) Solutions


Selina Solutions Concise Maths Class 10 Chapter 21 Trigonometrical Identities Exercise 21(D)

Question 1. Use tables to find sine of:

(i) 21°

(ii) 34° 42′

(iii) 47° 32′

(iv) 62° 57′

(v) 10° 20′ + 20° 45′

Solution:

(i) sin 21o = 0.3584

(ii) sin 34o 42’= 0.5693

(iii) sin 47o 32’= sin (47o 30′ + 2′) =0.7373 + 0.0004 = 0.7377

(iv) sin 62o 57′ = sin (62o 54′ + 3′) = 0.8902 + 0.0004 = 0.8906

(v) sin (10o 20′ + 2045′) = sin 30o65′ = sin 31o5′ = 0.5150 + 0.0012 = 0.5162

Question 2. Use tables to find cosine of:

(i) 2° 4’

(ii) 8° 12’

(iii) 26° 32’

(iv) 65° 41’

(v) 9° 23’ + 15° 54’

Solution:

(i) cos 2° 4’ = 0.9994 – 0.0001 = 0.9993

(ii) cos 8° 12’ = cos 0.9898

(iii) cos 26° 32’ = cos (26° 30’ + 2’) = 0.8949 – 0.0003 = 0.8946

(iv) cos 65° 41’ = cos (65° 36’ + 5’) = 0.4131 -0.0013 = 0.4118

(v) cos (9° 23’ + 15° 54’) = cos 24° 77’ = cos 25° 17’ = cos (25° 12’ + 5’) = 0.9048 – 0.0006 = 0.9042

Question 3. Use trigonometrical tables to find tangent of:

(i) 37°

(ii) 42° 18′

(iii) 17° 27′

Solution:

(i) tan 37= 0.7536

(ii) tan 4218′ = 0.9099

(iii) tan 17o  27′ = tan (1724′ + 3′) = 0.3134 + 0.0010 = 0.3144

Exercise 21(A) Solutions

Exercise 21(B) Solutions

Exercise 21(C) Solutions

Exercise 21(D) Solutions

Exercise 21(E) Solutions



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