Question

find the angle between the lines y=-rooot3x+5,y=1/root3x-2/root 3 ​

Answer 1

$\huge\pink{\boxed{\blue{\boxed{ \purple{ \boxed{{\pink{Answer}}}}}}}} \\ \large\pink{\boxed{\blue{\boxed{ \purple{ \boxed{{\pink{Your~answer↓}}}}}}}}$

Given :-

Two equation of lines

$1. \: y = \sqrt{3} + 5 \\ 2. \: y = \frac{1}{ \sqrt{3} } x - \frac{2}{ \sqrt{3} }$

To find :-

Angle between two lines.

Formula used :-

Let θ be the angle between two lines, then

$\boxed{ \large{ \mathfrak{tan θ= | \frac{m_1 - m_2}{1 + m_1 m_2} | }}}$

where

$m_1 \: = slope \: of \: ist \: line \\ m_2 \: = slope \: of \: 2nd \: line$

Solution:-

$\: y = \sqrt{3}x + 5 \\ \\slope \: of \: line \: m_1 = \sqrt{3} \\ \\ now \: \: y = \frac{1}{ \sqrt{3} } x - \frac{2}{ \sqrt{3} } \\ slope \: of \: line \: m_2 \: = \frac{1}{ \sqrt{3} }$

Using formula

$\boxed{ \large{ \mathfrak{tan θ= | \frac{m_1 - m_2}{1 + m_1 m_2} | }}}$

$tanθ = | \frac{ \sqrt{3} - \frac{1}{ \sqrt{3} } }{1 + \sqrt{3} \times \frac{1}{ \sqrt{3} } } | \\ = | \frac{3 - 1}{2 \sqrt{3} } | = \frac{2}{2 \sqrt{3} } \\ = | \frac{1}{ \sqrt{3} } | \\ = > θ = 30$

So angle between lines is 30° or 150°.

$\huge \fcolorbox{black}{cyan}{♛Hope it helps U♛}$

Answer 2

Explanation:

$\huge\pink{\boxed{\blue{\boxed{ \purple{ \boxed{{\pink{Answer}}}}}}}} \\ \large\pink{\boxed{\blue{\boxed{ \purple{ \boxed{{\pink{Your~answer↓}}}}}}}}$

Given :-

Two equation of lines

$1. \: y = \sqrt{3} + 5 \\ 2. \: y = \frac{1}{ \sqrt{3} } x - \frac{2}{ \sqrt{3} }$

To find :-

Angle between two lines.

Formula used :-

Let θ be the angle between two lines, then

$\boxed{ \large{ \mathfrak{tan θ= | \frac{m_1 - m_2}{1 + m_1 m_2} | }}}$

where

$m_1 \: = slope \: of \: ist \: line \\ m_2 \: = slope \: of \: 2nd \: line$

Solution:-

$\: y = \sqrt{3}x + 5 \\ \\slope \: of \: line \: m_1 = \sqrt{3} \\ \\ now \: \: y = \frac{1}{ \sqrt{3} } x - \frac{2}{ \sqrt{3} } \\ slope \: of \: line \: m_2 \: = \frac{1}{ \sqrt{3} }$

Using formula

$\boxed{ \large{ \mathfrak{tan θ= | \frac{m_1 - m_2}{1 + m_1 m_2} | }}}$

$tanθ = | \frac{ \sqrt{3} - \frac{1}{ \sqrt{3} } }{1 + \sqrt{3} \times \frac{1}{ \sqrt{3} } } | \\ = | \frac{3 - 1}{2 \sqrt{3} } | = \frac{2}{2 \sqrt{3} } \\ = | \frac{1}{ \sqrt{3} } | \\ = > θ = 30$

So angle between lines is 30° or 150°.

$\huge \fcolorbox{black}{cyan}{♛Hope it helps U♛}$