Given that sin πœƒ = a/b, then tan πœƒΒ  is equal to

(a) b/√(a 2 + b 2 )  (b) b/√(b 2 - a 2 )  (c) a/√(a 2 + b 2 )   (d) b/√(b^ 2 + a 2 )

Ques 26 (MCQ) - Given that sin ΞΈ = a/b, then tan ΞΈ is equal to [Video] - CBSE Class 10 Sample Paper for 2022 Boards - Maths Basic [MCQ]
part 2 - Question 26 - CBSE Class 10 Sample Paper for 2022 Boards - Maths Basic [MCQ] - Solutions of Sample Papers for Class 10 Boards - Class 10

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Question 26 Given that sin πœƒ = π‘Ž/𝑏, then tan πœƒ is equal to (a) 𝑏/√(π‘Ž^2 + 𝑏^2 ) (b) 𝑏/√(𝑏^2 βˆ’ π‘Ž^2 ) (c) π‘Ž/√(π‘Ž^2 + 𝑏^2 ) (d) 𝑏/√(𝑏^2 + π‘Ž^2 ) Let Ξ” ABC be the right angled triangle Given sin πœƒ = 𝒂/𝒃 Since sin πœƒ = π‘Άπ’‘π’‘π’π’”π’Šπ’•π’†/π‘―π’šπ’‘π’π’•π’†π’π’–π’”π’† Therefore, BC = a, AC = b Finding side AB By Pythagoras theorem AC2 = AB2 + BC2 b2 = AB2 + a2 b2 βˆ’ a2 = AB2 AB2 = b2 βˆ’ a2 AB = √(𝒃^πŸβˆ’π’‚^𝟐 ) Now, tan πœƒ = 𝐡𝐢/𝐴𝐡 = 𝒂/√(𝒃^𝟐 βˆ’ 𝒂^𝟐 ) So, the correct answer is (d)

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