# Thread: Application of similar triangles

1. ## Application of similar triangles

So I have this one problem:

A wall is being supported by a beam passing over a dumpster 5 feet high and 4 feet from the wall. Find the length of the shortest possible beam.

I know that you need to use similar triangles, but besides that I don't really know what to do. Please Help!!!!!!!!!
This is what I have figured out:

from the ground to the beam (along the wall) is v, and from the dumpster to the beam is v-5

from the wall to the beam (along the ground) is u, and from the dumpster to the beam is u-4.
the beam is x.

I've got it all into one equation: x^2 = v^2 + (20 +4)^2
(that's 20 over v-5) (v-5)

2. You can also do it with trig.

Let $\displaystyle {\theta}$ be the angle between the beam and the ground.

The length of the beam is $\displaystyle L=L_{1}+L_{2}$

$\displaystyle L_{1}$=length from ground to dumpster.

$\displaystyle L_{2}$= length from dumpster to wall.

$\displaystyle L=5csc({\theta})+4sec({\theta})$

$\displaystyle \frac{dL}{d{\theta}}=-5csc({\theta})cot({\theta})+4sec({\theta})tan({\th eta}), \;\ 0<{\theta}<\frac{\pi}{2}$

$\displaystyle =\frac{-5cos^{3}({\theta})+4sin^{3}({\theta})}{sin^{2}({\t heta})cos^{2}({\theta})}$

Now, $\displaystyle \frac{dL}{d{\theta}}=0$ if $\displaystyle 4sin^{3}({\theta})=5cos^{3}({\theta})$

Can you finish?.

3. Hello, jahichuanna!

This is a classic . . . The "shortest ladder" problem.

A wall is being supported by a beam passing over a dumpster 5 ft high and 4 ft from the wall.
Find the length of the shortest possible beam.
It is easier using Trigonometry . . .
Code:
                        * C
* |
*   |
B  * θ   |
* - - - * D
* |   4   |
*   |       |
*     |5      |
* θ     |       |
A * - - - - * - - - * E
F   4

In right triangle $\displaystyle BFA\!:\;\;\sin\theta \:=\:\frac{5}{AB} \quad\Rightarrow\quad AB \:=\:5\csc\theta$

In right triangle $\displaystyle CDB\!:\;\;\cos\theta \:=\:\frac{4}{BC} \quad\Rightarrow\quad BC \:=\:4\sec\theta$

So we have: .$\displaystyle AC \:=\:AB + BC \quad\Rightarrow\quad L \;=\;5\csc\theta + 4\sec\theta$

And that is the function we must minimize . . .

4. If you must use similar triangles, then:

y=height of wall and x=horizontal distance from ground to dumpster.

$\displaystyle \frac{5}{x}=\frac{y}{x+4}$

$\displaystyle y=\frac{5(x+4)}{x}$

Now, use Pythagoras and minimize.

You should get the same answer in both methods