two lines y=-2x+6 and y=1.5x+1 intersect. What is the size of the acute angle at the intersection?

Ive got 59.4 degrees. Any confirmation/slight alteration?

are you using the formula found here

http://planetmath.org/encyclopedia/AngleBetweenTwoLines.html ?

then the two slopes are 1.5 and -2

and tanß = │(-2-1.5)/(1+(-2)(1.5))│
tanß = 1.75
ß = 60.26º

To find the size of the acute angle at the intersection of two lines, you can use the slopes of the lines. Let's call the two lines Line 1 and Line 2, represented by the equations:

Line 1: y = -2x + 6
Line 2: y = 1.5x + 1

The slopes of Line 1 and Line 2 can be determined by comparing the coefficients of x in each equation. The slope-intercept form of a line is y = mx + c, where m is the slope of the line.

For Line 1, the slope is -2, and for Line 2, the slope is 1.5.

The acute angle between two lines is given by the formula:

angle = arctan(|(m2 - m1) / (1 + m1 * m2)|)

where m1 and m2 are the slopes of Line 1 and Line 2, respectively.

Plugging in the values, we get:

angle = arctan(|(1.5 - (-2)) / (1 + (-2) * 1.5)|)
= arctan(|3.5 / (-2.5)|)
= arctan(|-1.4|)

Now, let's calculate the value of the arctan(-1.4) in degrees:

angle = -55.15 degrees

Therefore, the size of the acute angle at the intersection of Line 1 and Line 2 is approximately 55.15 degrees, not 59.4 degrees.