Find the acute angle between the lines 2x-3y-6=0 and x+5y-20=0

Answer 1

2x-3y-6=0 has slope 2/3
x+5y-20=0 has slope -1/5

so, you want

arctan(-1/5) = -11.3°
arctan(2/3) = 33.7°

so, the acute angle between the lines is 45°

Answer 2

You should know that the slope of a line is the same as the tangent of the angle that the line makes with the x-axis

2x-3y-6=0 , slope = 2/3,
tanØ = 2/3,
Ø = appr 33.69° with the x-axis

same with x+5y-20=0
slope = -1/5 , line leans to the left.
angle with x-axis is appr 168.69°

angle between the lines = 168.69 - 33.69 = 135°

or, if Ø is the angle between them
tanØ = (m₂ - m₁) /( 1 + m₂m₁)
= (-1/5 - 2/3) / (1 + (-1/5)(2/3)
= (-13/15) / (13/15)
= -1
then Ø = arctan(-1) = 135°

Answer 3

Steve's answer of 45° is correct, I didn't see the "acute" angle part, and 180-135 = 45°