Question

find the angle between the liney=√3x and x-y=0​

Answer 1

Step-by-step explanation:

We know that y = mx + c so in line y=√3x

m = √3

And in line y=x

m=1

As tan x = m

x-1 = tan-¹(√3) = 60°

x-2 = tan-¹(1) = 45°

Angle = 60° - 45° = 15°

Second angle = 180° - 15° = 165

Answer 2

Answer:

The angle between the line y=√3x and x - y = 0 is ​15°

Step-by-step explanation:

Given:

The first line is y = √3x

We know that for the equation y = mx + c , the slope is m

Therefore, the slope = m₁ = √3

The second line is x - y = 0

Therefore, the slope = m₂ = 1

We know that tan $\theta$ = $\big| \frac{m_1-m_2}{1+m_1m_2}\big|$

where $\theta$ is the angle between two lines.

m₁ is the slope of the first line

m₂ is the slope of the second line

Therefore, on substitution,

=> tan  $\theta$ = $\big| \frac{\sqrt{3} -1}{1+\sqrt{3}*1}\big|$

=> tan  $\theta$ = $\big| \frac{\sqrt{3} -1}{1+\sqrt{3}}\big|$

=> tan  $\theta$ = $\big| \frac{\sqrt{3} -1}{1+\sqrt{3}}* \frac{1-\sqrt{3}}{1-\sqrt{3}}\big|$

=> tan  $\theta$ = $\big| \frac{\sqrt{3} -1-(\sqrt{3})^2 + \sqrt{3}}{1^2-(\sqrt{3})^2}\big|$

=> tan  $\theta$ = $\big| \frac{\sqrt{3} -1-3 + \sqrt{3}}{1-3}\big|$

=> tan  $\theta$ = $\big| \frac{2\sqrt{3} -4 }{-2}\big|$

=>  tan  $\theta$ = $\big| \frac{\sqrt{3} -2 }{-1}\big|$

=> tan  $\theta$ = $\big| 2-\sqrt{3} \big|$

=>  $\theta$ = $tan^{-1}\big| 2-\sqrt{3} \big|$

=>  $\theta$ = 15°

Therefore, the angle between the line y=√3x and x - y = 0 is ​15°