Trigonometry relations

1. Mar 11, 2009

Nyasha

1. The problem statement, all variables and given/known data

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)

3. 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

2. Mar 11, 2009

tiny-tim

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

3. Mar 11, 2009

Nyasha

What is OPA =π ?

4. Mar 11, 2009

tiny-tim

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

5. Mar 11, 2009

Nyasha

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

6. Mar 11, 2009

tiny-tim

just draw it … it's obvious!

7. Mar 11, 2009

Nyasha

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)

8. Mar 12, 2009

tiny-tim

waaa!

oh Nyasha …

you missed out poor little P!

9. Mar 12, 2009

Nyasha

Re: waaa!

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)

10. Mar 12, 2009

Looks good!