Question

The acute angle between intersecting lines that do not cross at right angles is the same as the angle determined by vectors normal to the lines or by the vectors parallel to the lines. Also, note that the vector ai + bj is perpendicular to the line ax + by = c Find the acute angles between the lines: x + √3y = 1 and (1 - √3)x + (1 + √3)y = 8 Thank you!

Answer 1

Answer:

$45^{\circ}$

Step-by-step explanation:

We are given that two lines equation

$x+\sqrt 3y=1$...(1)

$(1-\sqrt 3)y+(1+\sqrt 3)y=8$

Compare with the equation of line

ax+by+c=0

$a_1=1,b_1=\sqrt 3$

$a_2=(1-\sqrt 3),b_2=(1+\sqrt 3)$

The angle between two lines =Angle between two vectors

The angle between two vector

$a_1i+b_1j$ and

$a_2i+b_2j$

is given by

$cos\theta=\frac{a_1a_2+b_1b_2}{\sqrt{a^2_1+b^2_1}\sqrt{a^2_2+b^2_2}}$

Using the formula

Therefore, the angle between two lines

$cos\theta=\frac{1(1-\sqrt 3)+\sqrt 3(1+\sqrt 3)}{\sqrt{(1)+(\sqrt 3)^2}\times \sqrt{(1-\sqrt 3)^2+(1+\sqrt 3)^2}}$

$cos\theta=\frac{1-\sqrt 3+\sqrt 3+3}{\sqrt{1+3}\times\sqrt{1+3-2\sqrt 3+1+3+2\sqrt 3}}$

$cos\theta=\frac{4}{2\times\sqrt 8}=\frac{2}{2\sqrt 2}$

$cos\theta=\frac{1}{\sqrt 2}$

$cos\theta=cos45^{\circ}$

By using $cos45^{\circ}=\frac{1}{\sqrt 2}$

$\theta=45^{\circ}$

Hence, the angle between two lines =45 degree