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

## 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

tanθ

_{1}= √5, tanA = √(77)/2 (since cosA = 2/9, and A is an acute angle.)tanθ

_{2}= m, where m is the slope to be found.Since θ

_{2}-θ_{1 }= ±A,_{ }m = tan(θ

_{1}±A) = [tanθ_{1}±tanA]/[1-+tanθ_{1*}tanA]Plugging in tanθ

_{1}= √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

m

_{1}= tan B_{1}= √5m

_{2}= tan B_{2}= ?The angle A between the two lines can be

A = B

_{2}- B_{1}or

A = B

_{1}- B_{2}depending on which line is steeper.

Therefore,

B

_{2}= B_{1}+ Aor

B

_{2}= B_{1}- AIn your case, B

_{1}=tan^{-1}√5=65.9° and A = cos^{-1}(2/9)=77.2°, so thatB

_{2}= tan^{-1}√5 ± cos^{-1}(2/9)B

_{2}= 143°=-36.9°or

B

_{2}= -11.3°The slopes are

tan(-36.9)=-0.75=-3/4

and

tan(-11.3)=-0.2=-1/5