what acute angle does a line of slope -2/3 make with vertical line.
To find the acute angle that a line with a slope of -2/3 makes with a vertical line, we can use the property that the slopes of perpendicular lines are negative reciprocals of each other.
Since the given line has a slope of -2/3, the perpendicular line will have a slope of 3/2.
Now, we know that a vertical line has an undefined slope, so the slope of the vertical line can be represented as infinity (∞).
To find the acute angle between these two lines, we can use the formula:
tan(angle) = |(m1 - m2) / (1 + m1 * m2)|
where m1 and m2 are the slopes of the two lines.
Substituting the values, we have:
tan(angle) = |((-2/3) - (3/2)) / (1 + (-2/3) * (3/2))|
Calculating this expression:
tan(angle) = |((-2/3) - (9/6)) / (1 + (-6/6))|
tan(angle) = |((-4/6) - (9/6)) / (6/6 - 6/6)|
tan(angle) = |(-13/6) / 0|
Since the denominator is 0, the angle is 90 degrees (π/2 radians).
Therefore, the acute angle between the line with slope -2/3 and the vertical line is 90 degrees (π/2 radians).
To find the acute angle between a line with slope -2/3 and a vertical line, we can use the fact that the slope of a line is the tangent of its angle with the x-axis.
First, let's find the slope of the vertical line. Since a vertical line is perfectly perpendicular to the x-axis, its slope is undefined.
Now, let's find the slope of the line with the given slope -2/3. We'll call this slope m1.
m1 = -2/3
Next, let's find the angle that m1 makes with the x-axis (which is also the acute angle it makes with the vertical line). We'll call this angle θ.
tan(θ) = m1
Since we know the value of m1, we can find the value of θ by taking the inverse tangent (arctan) of m1.
θ = arctan(m1)
θ = arctan(-2/3)
Using a calculator, we find:
θ ≈ -33.69 degrees
Since we are interested in the acute angle, we can take the absolute value of θ to make it positive:
|θ| ≈ 33.69 degrees
Therefore, the acute angle between the line with slope -2/3 and the vertical line is approximately 33.69 degrees.