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

(A) b/√(b^2  - a^2 )   

(B) b/a  

(C) √(b^2  - a^2 )/b   

(D) a/√(b^2  - a^2 )





  1. Chapter 8 Class 10 Introduction to Trignometry (Term 1)
  2. Serial order wise


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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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.