# Example 15

Last updated at March 9, 2017 by Teachoo

Last updated at March 9, 2017 by Teachoo

Transcript

Example 15 Find the angle between the lines y − √3x − 5 = 0 and √3y − x + 6 = 0 . Let the lines be y − √3 x − 5 = 0 √3 y − x + 6 = 0 We know that angle between 2 lines (θ) can be found by using formula tan θ =|(𝑚_2 − 𝑚_1)/(1 + 𝑚_2 𝑚_1 )| Let the slope of line (1) be m1 & slope of line (2) be m2 Angle between two lines is given tan θ =|(𝑚_2 − 𝑚_1)/(1 + 𝑚_2 𝑚_1 )| Putting values tan θ = |(1/√3 − √3)/(1 + 1/√3 × √3)| = |(1/√3 − √3)/(1 + 1)| = |((1 − (√3)(√3))/√3)/2| = |(1 − 3)/(2√3)| = |( − 2)/(2√3)| = |( − 1)/√3| = 1/√3 Thus, tan θ = 1/√3 We know that tan 30° = 1/√3 So, tan θ = tan 30° θ = 30° Thus, the acute angle between the lines (1) & (2) is θ = 30° & obtuse angle between (say ϕ) these two lines is ϕ = 180 − θ ϕ = 180 − 30° ϕ = 150° Thus, the required angle between lines is 30° or 150° Note There are always two angles between the lines, one acute angle θ & other obtuse angle ϕ which are in linear pair, Thus θ + ϕ = 180° ϕ = 180° – θ

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Example 15 Important You are here

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Example 22 Important

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Class 11

Important Question for exams Class 11

- Chapter 1 Class 11 Sets
- Chapter 2 Class 11 Relations and Functions
- Chapter 3 Class 11 Trigonometric Functions
- Chapter 4 Class 11 Mathematical Induction
- Chapter 5 Class 11 Complex Numbers
- Chapter 6 Class 11 Linear Inequalities
- Chapter 7 Class 11 Permutations and Combinations
- Chapter 8 Class 11 Binomial Theorem
- Chapter 9 Class 11 Sequences and Series
- Chapter 10 Class 11 Straight Lines
- Chapter 11 Class 11 Conic Sections
- Chapter 12 Class 11 Introduction to Three Dimensional Geometry
- Chapter 13 Class 11 Limits and Derivatives
- Chapter 14 Class 11 Mathematical Reasoning
- Chapter 15 Class 11 Statistics
- Chapter 16 Class 11 Probability

About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .