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lines a(1) and a(2 in two-dimensional space intersect at a point, forming an acute angle A

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2 Answers

tanθ1 = √5, tanA = √(77)/2 (since cosA = 2/9, and A is an acute angle.)
 
tanθ2 = m, where m is the slope to be found.
Since θ21 = ±A,
m = tan(θ1±A) = [tanθ1±tanA]/[1-+tanθ1*tanA]
Plugging in tanθ1 = √5, tanA = √(77)/2 gives you the exact final answers.
 
Answer in decimal form: m = -.199 or m = -.752.
The relation between the slope of a line and the angle B it makes with the horizontal (x-axis) is
m = tan B.
 
You have two lines with slopes
m1 = tan B1 = √5
m2 = tan B2 = ?
 
The angle A between the two lines can be
 
A = B2 - B1
or
A = B1 - B2
depending on which line is steeper.
Therefore,
B2 = B1 + A
or
B2 = B1 - A
 
In your case, B1=tan-1√5=65.9° and A = cos-1(2/9)=77.2°, so that
 
B2 = tan-1√5 ± cos-1(2/9)
B2 = 143°=-36.9°
or
B2 = -11.3°
The slopes are
tan(-36.9)=-0.75=-3/4
and
tan(-11.3)=-0.2=-1/5

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