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

Question

lines a(1) and a(2 in two-dimensional space intersect at a point, forming an acute angle A. the slope of a(1) is square root of 5 , and cos A= 2/9. What is the slope of a(2)? There are two answers, both must be given to properly answer the question

Answer 1

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 θ₂-θ₁ = ±A,

m = tan(θ₁±A) = [tanθ₁±tanA]/[1-+tanθ_1*tanA]

Plugging in tanθ₁ = √5, tanA = √(77)/2 gives you the exact final answers.

 

Answer in decimal form: m = -.199 or m = -.752.

Answer 2

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