Answer:
[tex]\theta=45^{\circ}[/tex]
Step-by-step explanation:
We are given that the equation of lines
[tex]x+\sqrt 3y=1[/tex]
[tex](1-\sqrt 3)x+(1+\sqrt 3)y=8[/tex]
According to question
The vector perpendicular to the lines is given by
[tex]i+\sqrt 3j[/tex] and [tex](1-\sqrt 3)i+(1+\sqrt 3)j[/tex]
Therefore, the angle between two vectors is given by
[tex]cos\theta=\frac{a_1a_2+b_1b_2}{\sqrt{a^2_1+b^2_1}\sqrt{a^2_2+b^2_2}}[/tex]
Using the formula
[tex]cos\theta=\frac{1(1-\sqrt 3)+\sqrt 3(1+\sqrt 3)}{2\times 2\sqrt 2}[/tex]
[tex]cos\theta=\frac{1-\sqrt 3+\sqrt 3+3}{4\sqrt 2}=\frac{1}{\sqrt 2}[/tex]
[tex]cos\theta=cos 45^{\circ}[/tex]
[tex]\theta=45^{\circ}[/tex]
Hence, the acute angle between the lines is given by
[tex]\theta=45^{\circ}[/tex]
