Given that sin ฮธ = a/b, then cos ฮธ is equal to

(A) b/โˆš(b^2ย  - a^2 ) ย ย 

(B) b/a ย 

(C) โˆš(b^2ย  - a^2 )/b ย ย 

(D) a/โˆš(b^2ย  - a^2 )

MCQ - Class 10 Boards - Given that sin ฮธ = a/b, then cos ฮธ is equal - NCERT Exemplar - MCQ

part 2 - Question 7 - NCERT Exemplar - MCQ - Serial order wise - Chapter 8 Class 10 Introduction to Trignometry

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Transcript

Question 7 Given that sin ฮธ = ๐‘Ž/๐‘, then cos ฮธ is equal to (A) ๐‘/โˆš(๐‘^2 โˆ’ ๐‘Ž^2 ) (B) ๐‘/๐‘Ž (C) โˆš(๐‘^2 โˆ’ ๐‘Ž^2 )/๐‘ (D) ๐‘Ž/โˆš(๐‘^2 โˆ’ ๐‘Ž^2 ) Now, cos2 ฮธ = 1 โˆ’ sin2 ฮธ Putting sin ฮธ = ๐‘Ž/๐‘ cos2 ฮธ = 1โˆ’(๐‘Ž/๐‘)^2 cos2 ฮธ = 1โˆ’๐‘Ž^2/๐‘^2 cos2 ฮธ = (๐’ƒ^๐Ÿ โˆ’ ๐’‚^๐Ÿ)/๐’ƒ^๐Ÿ Taking square root cos ฮธ = ยฑ โˆš((๐‘^2 โˆ’ ๐‘Ž^2)/๐‘^2 ) cos ฮธ = ยฑ โˆš(๐‘^2 โˆ’ ๐‘Ž^2 )/๐‘ Since only positive values are given in options cos ฮธ = โˆš(๐’ƒ^๐Ÿ โˆ’ ๐’‚^๐Ÿ )/๐’ƒ So, the correct answer is (C)

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