help! can you do this math101 question?

[s gonzales]

A right triangle has base b and height h1. An infinite sequence of nested right tringles is constructed inside the triangle by dropping perpendiculars as shown on the image. a) Find and expression for the edge hn. b) Find an expression for the infinite sum: lim?→∞ (ℎ1 + ℎ2 + ⋯ + ℎ? )

very important for me.

 

[Dr.Peterson]

Where are you stuck?

If it is on the first part, you might first find the ratio [MATH]h_1:h_2[/MATH]. Is [MATH]h_2:h_3[/MATH] the same?

In any case, please show what you have done (or even just tried) so we can see what help you need.

 

[ksdhart2]

Just in the interest of full disclosure, this exact problem was also posted on the AskMath subreddit, presumably by the same person because the usernames are nearly identical. There, the student was given the following hint, but has yet to respond:

Dropping the line h₂ creates a pair of triangles similar to the one you started with.

 

[Jagulba]

Find the angle theta in each of the new triangles.
You will see that cos(theta) will give you the relationship between two consecutive h ( hn and hn+1). for h1 you can write it down as function of b and theta.
The hint from me would be that the x in the hint is cos(theta)

and thanks that was fun to do

 

[s gonzales]

what do you think?

 

[s gonzales]

is this correct and enough for the answer?

 

[MarkFL]

As alluded to above, we see that:

[MATH]h_{n}-h_{n-1}\cos(\theta)=0[/MATH]
This gives us a characteristic root of:

[MATH]r=\cos(\theta)[/MATH]
And so the homogeneous solution is:

[MATH]h_n=c_1\cos^n(\theta)[/MATH]
To determine the parameter, we observe that:

[MATH]h_1=c_1\cos(\theta)\implies c_1=\frac{h_1}{\cos(\theta)}[/MATH]
Hence:

[MATH]h_n=h_1\cos^{n-1}(\theta)[/MATH]
Now for the partial sum \(S_n\):

[MATH]S_n=h_1\sum_{k=1}^{n}\left(\cos^{k-1}(\theta)\right)=h_1\sum_{k=0}^{n-1}\left(\cos^{k}(\theta)\right)[/MATH]
Applying the hint given for this sum, what do you get?

 

[MarkFL]

To follow up:

[MATH]S_n=h_1\frac{\cos^n(\theta)-1}{\cos(\theta)-1}[/MATH]
Hence:

[MATH]S_{\infty}=\lim_{n\to \infty}S_n=\frac{h_1}{1-\cos(\theta)}[/MATH]