Geometry, or Trig?

[Lokito]

I am currently in a pre-calculus class at school. The problem below came from a "Right Triangle Trig Challenge Problems" worksheet. I solved all the other problems, but cannot figure this one out. The teacher said he originally thought it was a trig problem, but later realized it was just geometry. I think he may have been mistaken. Either way, I was never good at geometry, so it's not a surprise I can't figure it out. The problem is graphical, so I attempted to recreate it using Paint, thinking it'd be clearer than my explanation. I didn't see anything against posting links in the forum rules, so if you will, please check out the imageshack link below to see the problem:

http://img152.imageshack.us/img152/2419/trigms9.jpg

Ignore the incorrect scale, please. I don't know how to create an image to scale on a computer. Also, measurements are in feet.

Now, on to what I've done:

Using the knowledge that it's a 3-4-5 triangle, I labeled one of the diagonals as 50 ft. Next, I found all the angles in the diagram. I also split the triangle with hypotenuse of 40 into two right triangles for clarification. I found the lower (after intersection) segment of the larger diagonal using:
sin(53)=opposite/30
opposite=30sin(53)
Approx. = 23.959

BUT, using Pythagorean theorem I found the length of the larger diagonal to be the square root of 8000, or about 89.443.
When I use sin(53)=opposite/80
opposite=80sin(53)
Approx. = 63.891

89.443 - 63.891 doesn't equal anything close to 23.959. It equals about 25.552
Wait - I just saw that I used that second sine equation on a non-right triangle. We haven't learned that yet. Probably why it's wrong.

I don't know how to solve this. Any help would be appreciated.

 

[stapel]

You have:

   |\
   | \
   |  \
80 |   \ *|
   |   *\ | 30
   | *   \| 
   --------
      40
(very much NOT to scale!!)

...and you are needing to find the height of the point where the two slanty sides intersect.

Try using algebra. Find the equations of the two lines containing by the slanty segments, and then find where they intersect. The y-value of their intersection point will be the height of the intersection point.

Eliz.

 

[jwpaine]

Like stable said, find an equation for each line, and set them equal to each other.

line_1 = line_2
m_1(x) + b_1 = m_2(x) + b_2

Solve for x, and then plug the value of x back into either line equation to get the value of y (vertical distance)

 

[Lokito]

Like stable said, find an equation for each line, and set them equal to each other.

line_1 = line_2
m_1(x) + b_1 = m_2(x) + b_2

Solve for x, and then plug the value of x back into either line equation to get the value of y (vertical distance)

I don't know why I didn't think of that in the first place. Here's the algebraic solution, just for myself (have to think I can do some math, right?):
.75x = -2x + 80
.75x = 80 + -2x
.75x + 2x = 80 + -2x + 2x
2.75x = 80
x = 29.09090909

.75 * ANS = 21.8181...

Here's another question: should I round the x value before solving for y?

 

[jwpaine]

Like stable said, find an equation for each line, and set them equal to each other.

line_1 = line_2
m_1(x) + b_1 = m_2(x) + b_2

Solve for x, and then plug the value of x back into either line equation to get the value of y (vertical distance)

I don't know why I didn't think of that in the first place. Here's the algebraic solution, just for myself (have to think I can do some math, right?):
.75x = -2x + 80
.75x = 80 + -2x
.75x + 2x = 80 + -2x + 2x
2.75x = 80
x = 29.09090909

.75 * ANS = 21.8181...

Here's another question: should I round the x value before solving for y?

Very good.

I prefer to keep my work in fractional form to minimize error. Only at the end will I get a decimal approximation. If you get a decimal approximation every step, then you could potentially have an increasing % error.

-2x + 80 = (3/4)x
-8x + 320 = 3x
x = 320/11
y = -2(320/11) + 80 = [240/11]

 

[Lokito]

Oops, my original message of thanks never went through. So, thanks!