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ladykimchi · (This post was last modified: 06-25-2010 03:49 PM by ladykimchi.)

Solve the following,in which C=90*

A ladder 42 feet long is placed so that it will reach a window 30 feet high on one side of a street; if it is turned over,its foot being held in position it will reach a window 25 feet high on the other side of the street. How wide is the street from building to building?

the picture attached was my illustration of it

**eigrihs123** · 06-25-2010, 03:50 PM

Hello ladykimchi,

I'm somebody from plurk. You will sure get a good help from forum members here. But for now, I can help you with your assignment. Just give me 15 minutes and I'll post the answer here.

ladykimchi · 06-25-2010, 03:55 PM

thank you so much for helping me!

**eigrihs123** · 06-25-2010, 04:13 PM

Hello Ladykimchi...does the illustration you attached has something to do with the problem? Or is it another problem? It looks like it does not coincide with the illustration in the problem.

ladykimchi · (This post was last modified: 06-25-2010 07:52 PM by zencontrol.)

hala,sa isang problem pala un,sorry

Translated: Sorry, it should be for another question

**eigrihs123** · (This post was last modified: 06-25-2010 04:31 PM by eigrihs123.)

its ok. I am just wondering though...lol....I'll continue with my own illustration

Ok, this is my answer.

After analyzing the proble, I have come up with a simple illustration of the problem on the attached image.

This is how to solve it:

Let A= distance from one building to another building
42= height of the ladder
25= height of the window that can be reached by a ladder from the other side of the street

Sin θ = 25/ 42

θ = arcsin (25/42)

θ = 36. 53 degrees

solving for A:

tan θ = 25/ A

A= tan θ * 25
A = tan 36.53 * 25

A = 33.75 feet

What do you think?

ladykimchi · 06-25-2010, 04:39 PM

thank you for helping me!,i gotta go,see you later!

***zencontrol*** · 06-25-2010, 07:53 PM

Nice to see a thread for homework. This could be interesting for all students.

arkj75 · 06-26-2010, 08:47 PM

How many problems do we have here. I think I am confused?!?!?

**eigrihs123** · 06-26-2010, 08:49 PM

(06-26-2010 08:47 PM)arkj75 Wrote: How many problems do we have here. I think I am confused?!?!?

The question has been mixed. The problem does not coincide with the image. Anyway, I have been able to answer it, you may check it and find out if it is correct?