Question

The measure of the acute angle between lines √3x-y-2=0 and x-√3y+1=0​

Answer 1

let √3x - y - 2 = 0 be line 1

and x - √3y + 1 = 0 be line 2

${m}_{1} = \frac{- (\sqrt{3})}{- 1} = \sqrt{3} \\</p><p> m_{2} = \frac{- 1}{- \sqrt{2}} = \frac{1}{\sqrt{3}}$

... ($m = \frac{-a}{b}$)

now, for $\alpha$ as the angle between the two given lines, we know that;

$tan \alpha \\ = | \frac{ m_{1} - m_{2} }{1 + m_{1} m_{2} } | \\ = | \frac{ \sqrt{3} - \frac{1}{ \sqrt{3} } }{1 + \sqrt{3} \times \frac{1}{ \sqrt{3} } } | = | \frac{ \frac{3 - 1}{ \sqrt{3} } }{1 + 1} | = | \frac{2}{2 \sqrt{3} } | \\ = | \frac{1}{ \sqrt{3} } |$

$= ±\frac{1}{\sqrt{3}} \\</p><p>\alpha = {tan}^{- 1}(±\frac{1}{\sqrt{3}}) \\ </p><p>\alpha = {30}^{o} or {150}^{o}$

since we're supposed to find the acute angle, the angle between them is 30°

OR

Since slopes of the lines are √3 and 1/√3

their inclinations with x-axis are 60° and 30° respectively. Hence, acute angle between them is 30°.