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NDA (Held On: 17 April 2016) Maths Previous Year paper

Option 1 : 30°

__Concept:__

**Angle between two lines:** The angle θ between the lines having slope m_{1} and m_{2} is given by

\(\tan {\rm{\theta }} = {\rm{\;}}\left| {\frac{{{{\rm{m}}_2} - {{\rm{m}}_1}}}{{1 + {\rm{\;}}{{\rm{m}}_1}{\rm{\;}}{{\rm{m}}_2}}}} \right|\)

__Calculation:__

Given: y - √3x – 5 = 0 & √3y – x + 6 = 0

y - √3x – 5 = 0

⇒ y = √3x + 5

So, slope of line, m_{1} = √3

√3y – x + 6 = 0

\(\Rightarrow {\rm{y}} = \frac{{\rm{x}}}{{\sqrt 3 }} - \frac{6}{{\sqrt 3 }}\)

So, slope of the line, m_{2} = \(\frac{1}{{\sqrt 3 }}\)

Let θ be the acute angle between the lines.

\(\tan {\rm{\theta }} = \left| {\frac{{{{\rm{m}}_1} - {{\rm{m}}_2}}}{{1 + {{\rm{m}}_1}{{\rm{m}}_2}}}} \right|\)

\(\Rightarrow \tan {\rm{\theta }} = \left| {\frac{{\sqrt 3 - \frac{1}{{\sqrt 3 }}}}{{1 + \sqrt 3 \times \frac{1}{{\sqrt 3 }}}}} \right|\)

\( \Rightarrow \tan {\rm{\theta }} = \left| {\frac{{\frac{2}{{\sqrt 3 }}}}{2}} \right|\)

\(\Rightarrow \tan {\rm{\theta }} = \frac{1}{{\sqrt 3 }}\)

⇒ θ = 30°

Electric charges and coulomb's law (Basic)

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