Given that sin πœƒ = a/b, then tan πœƒ  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 )

Slide45.jpeg

Advertisement

Slide46.jpeg

Advertisement

  1. Class 10
  2. Solutions of Sample Papers for Class 10 Boards

Transcript

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)

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
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.