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

(A) b/√(b^2  - a^2 )   

(B) b/a Β 

(C) √(b^2  - a^2 )/b   

(D) a/√(b^2  - a^2 )



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Question 7 Given that sin ΞΈ = π‘Ž/𝑏, then cos ΞΈ is equal to (A) 𝑏/√(𝑏^2 βˆ’ π‘Ž^2 ) (B) 𝑏/π‘Ž (C) √(𝑏^2 βˆ’ π‘Ž^2 )/𝑏 (D) π‘Ž/√(𝑏^2 βˆ’ π‘Ž^2 ) Now, cos2 ΞΈ = 1 βˆ’ sin2 ΞΈ Putting sin ΞΈ = π‘Ž/𝑏 cos2 ΞΈ = 1βˆ’(π‘Ž/𝑏)^2 cos2 ΞΈ = 1βˆ’π‘Ž^2/𝑏^2 cos2 ΞΈ = (𝒃^𝟐 βˆ’ 𝒂^𝟐)/𝒃^𝟐 Taking square root cos ΞΈ = Β± √((𝑏^2 βˆ’ π‘Ž^2)/𝑏^2 ) cos ΞΈ = Β± √(𝑏^2 βˆ’ π‘Ž^2 )/𝑏 Since only positive values are given in options cos ΞΈ = √(𝒃^𝟐 βˆ’ 𝒂^𝟐 )/𝒃 So, the correct answer is (C)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.