Check sibling questions

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.