Question

find the angle between y=-√3x+5,y=1/√3x-2/√3​

Answer 1

Step-by-step explanation:

y = mx + c ( where m is slope)

slope → first line,

y-√3x - 5 = 0

y = √3x + c => m1 =√3 and c1=5

similarly, slope → second line

√3y-x+6=0

y= (x/√3) -(6/√3) => m2= 1/√3 and c2=6/√3 = 6√3/3=2√3

Angle between 2 lines is given by:-

tanθ = (m1-m2)/(1+m1m2)

=(√3–1/√3)/(1+√3*1/√3)

=(2/√3)/2

=1/√3

tanθ = 1/√3

=>θ =30∘