Angle Between Two Lines

[nycmathguy]

Section 5.4

I need help with 97.

 

[MathLover1]

97.

The angle θ between the lines having slope m1 and m2 is given by

tan θ = (m2−m1)/(1+m1*m2)

y=x -> slope m1=1
y=sqrt(3)x-> slope m2=sqrt(3)

tan θ = (sqrt(3)-1)/(1+1*sqrt(3))

tan θ = (sqrt(3)-1)/(1+sqrt(3)) ......rationalize

tan θ = 2 - sqrt(3)

θ =tan^-1(2 - sqrt(3))

θ =π/12 (result in radians)

θ =15°

 

[nycmathguy]

97.

The angle θ between the lines having slope m1 and m2 is given by

tan θ = (m2−m1)/(1+m1*m2)

y=x -> slope m1=1
y=sqrt(3)x-> slope m2=sqrt(3)

tan θ = (sqrt(3)-1)/(1+1*sqrt(3))

tan θ = (sqrt(3)-1)/(1+sqrt(3)) ......rationalize

tan θ = 2 - sqrt(3)

θ =tan^-1(2 - sqrt(3))

θ =π/12 (result in radians)

θ =15°

Explain this:

y=x -> slope m1=1
y=sqrt(3)x-> slope m2=sqrt(3)

 

[MathLover1]

y=x -> compare to slope intercept form y=mx+b=> slope m1=1
y=sqrt(3)x-> compare to slope intercept form y=mx+b=> slope m2=sqrt(3)

 

[nycmathguy]

y=x -> compare to slope intercept form y=mx+b=> slope m1=1
y=sqrt(3)x-> compare to slope intercept form y=mx+b=> slope m2=sqrt(3)

Compare to slope intercept form y = mx + b in what way?

 

[nycmathguy]

97.

The angle θ between the lines having slope m1 and m2 is given by

tan θ = (m2−m1)/(1+m1*m2)

y=x -> slope m1=1
y=sqrt(3)x-> slope m2=sqrt(3)

tan θ = (sqrt(3)-1)/(1+1*sqrt(3))

tan θ = (sqrt(3)-1)/(1+sqrt(3)) ......rationalize

tan θ = 2 - sqrt(3)

θ =tan^-1(2 - sqrt(3))

θ =π/12 (result in radians)

θ =15°

I would like you to provide steps when answering my questions. Steps help to clarify what is happening as we attempt to find the right answer.

Sample:

Solve 3x = 15 for x.

Steps:

1. Divide both sides by 3, the coefficient of x.

3x/3 = 15/3

x = 5

2. We can also multiply both sides by the reciprocal of the coefficient of x.

The reciprocal of 3 is 1/3.

(1/3)(3x) = (1/3)(15)

x = 5

Either way, we find the right answer which is 5.
Mira, can you do this with precalculus and beyond?
Can you provide steps for precalculus and beyond when answering questions?

 

[MathLover1]

I do not think you need every single step to be provided, you are not beginner
but, if you insist, I can do it

 

[MathLover1]

Compare to slope intercept form y = mx + b in what way?

if you are given line y=x, you can write it as

y=1*x+0 then you compare to slope intercept form
y = mx + b
so, m=1 , b=0

 

[nycmathguy]

if you are given line y=x, you can write it as

y=1*x+0 then you compare to slope intercept form
y = mx + b
so, m=1 , b=0

Ok. Now I get it.