Finding the height of my two vertical posts

[jwpaine]

I need to find the height of each one of my red, vertical support posts.

This is the only data I have for the project I have done.... I have gotten to this point, and need some help.

Using the distance formula between the x intercept of my purple line, and the black dot, is approx 2.6"

I then used trig to find the angle:

Sin(x) = (25/36)/2.6
Sin^(-1)((25/36)/2.6) = 15.50 degrees for angle of purple line

How on earth can I find the height of each of the red support struts? My mirror is 1" in length between each.


This is not school related, I am making a parabolic reflector using plane segments. This is the last part I need help on and then I can begin construction. (the purple line is actually tangent to (1/36)x^2 at the point shown on the diagram. I erased everything not pertaining to this post.

Thanks for the help.

 

[jonboy]

How about using the distance formula:\(\displaystyle \L \;d\,=\,sqrt{(x_1\,-\,x_2)^2\,+\,(y_1\,-\,y_2)^2}\)

Since we know the point on the x - axis is 4.5 (if your diagram is drawn to scale) then of course solve:

\(\displaystyle \L \;y\,=\,\frac{5}{18}\,\cdot\,\frac{9}{2}\,-\,\frac{25}{36}\)

So you have already:\(\displaystyle \L \;(0\,,\,\frac{9}{2})\)

Now you'll have\(\displaystyle \L \;(\frac{9}{2}\,,\,y)\)

Sorry if I'm assuming too much, just trying to help my fellow math cohort.

I get approx 5 units.

 

[jwpaine]

I wish it was that easy...but its not drawn to scale... it would be if my mirror was exactly one inch in proportion with everything else, but its not. If it were...then I would simply find the distance from 0 to 2.6-(1/2)(mirror-length) = 2.1" for the first strut.

I just drew that mirror at 1" as a representation of its length with a label....infact, it may be in proportion.....

Thanks for the help, though.

 

[jwpaine]

Here is a simplified diagram with all the data included that I have calculated:

I need to find the coordinates for points B and D

Thanks for any and all help, I really appreciate it.
John P.

 

[Denis]

John, draw verticals BE, CF and DG (E, F and G on x axis).
Then draw horizontals BH (H on CF) and CI (I on DG).
That'll give tyou 2 congruent right triangles: BCH and CDI.
And these are similar to right triangle ACF : OK????

 

[jwpaine]

So for point B:

AC/CF = BC/CH
2.63/(25/36) = 0.5/CH CH = .132024
Point B y = (25/36)-.132024 = 0.56242

AC/AF = BC/BH
2.63/5 = 0.5/BH BH = .95057
Point B x = 5-.95057 =4.04943

Point B = (4.04943, 0.56242)

X coordinate for B is correct.. Y coordinate isn't. I have checked algebraically and graphically, and my solution doesn't work.

What have I done wrong? I'm starting to get frustrated by this.
Thanks.
John.

EDIT: Substituting X into my liner equation and solving for y works.
((5/18)(4.03648) - (25/36)) y = 0.4268

 

[Denis]

That's probably(!) because sqrt[(5/2)^2 + (25/36)^2] = 2.5946... : NOT 2.63 :idea:

Btw, sqrt[(5/2)^2 + (25/36)^2] simplifies to sqrt(8725 / 1296)