Trigonometric Relations for Line Segments with Fixed and Movable Points

[Nyasha]

Homework Statement

Straight line segments are drawn from the fixed point P1(0,1) and P2(3,2) to the movable point P, with coordinates (x,0)on the positive x-axis.

Assuming that 0 ≤ x ≤ 3, show that the angle θ between the two line segments PP1 and PP2 is given by the relation:

θ= π-arccotx-arccot(3-x/2)

The Attempt at a Solution

tan\theta = (3-0)/(2 - 1) = 3/2 For PP1

\tan\theta = (3-x)/(2-0)= (3 - x)/2 For PP2

\theta = arctan(3/2) - arctan((3-x)/2)

First of all l do not know if l am on the right path and even if l am on the right path l get stuck here. Please help

 

[tiny-tim]

Hi Nyasha!

tan\theta = (3-0)/(2 - 1) = 3/2 For PP1

No, that's the line from P1 to P2, not P to P1

First of all l do not know if l am on the right path

I'm not sure either … it's difficult to tell whether you're just guessing …

to make it clear to the examiner that you're not guessing, give the angles or directions names, so you can make it clear which angles you're subtracting from which, and why …

in this case, call (0,0) O, and call (3,0) A, and use OPA = π

 

[Nyasha]

Hi Nyasha!


No, that's the line from P1 to P2, not P to P1


I'm not sure either … it's difficult to tell whether you're just guessing …

to make it clear to the examiner that you're not guessing, give the angles or directions names, so you can make it clear which angles you're subtracting from which, and why …

in this case, call (0,0) O, and call (3,0) A, and use OPA = π

What is OPA =π ?

 

[tiny-tim]

What is OPA =π ?

uhh? OPA is the angle OPA, from (0,0) to P to (3,0)

 

[Nyasha]

uhh? OPA is the angle OPA, from (0,0) to P to (3,0)

Okay, so any tips on how l prove that θ= π-arccotx-arccot(3-x/2) ? I am really confused/blank on this question.

 

[tiny-tim]

just draw it … it's obvious!

 

[Nyasha]

just draw it … it's obvious!

Tim l have drawn it but it is still not obvious. I still can't figure out how they came up with the relation θ= π-arccotx-arccot(3-x/2)

 

[tiny-tim]

waaa!

oh Nyasha …

you missed out poor little P! ​

 

[Nyasha]

oh Nyasha …

you missed out poor little P! ​

Tim l think l solved it now, l drew a better diagram and then figure out what to do. Here is what l did:


\theta= ArcCot((x-0)/(0-1)) = ArcCot(-x) = Pi - ArcCot(x) This one is the slope for PP1

\theta = ArcCot((3-x)/(2-0) = ArcCot((3-x)/2)

\theta = Pi - ArcCot(x) - ArcCot((3-x)/2) (subtract PP1 from PP2)

 

[tiny-tim]

Looks good!