A ladder that is 4 ft long is leaning against a wall. Under the ladder ( up against the wall, on the floor) sits a cubical box with side length 1 ft. The ladder is simultaneously touching the edge of the box, the wall, and the floor.
What is the vertical distance from the top of the box to the top of the ladder?
This is the best I could do for a diagram. I think its sufficient to get the idea, if not let me know.
@joe-shmosaid This is a little tricky algebra... certainly its more involved than one might initially suspect.
A solution is still up for grabs!
I was never really interested in Trigonometry,but out of curiosity ,I looked on the net to see if it was possible to find angles in right angled triangles when given just one side measurement and it talked about something called Pythagorean triples with equations (m² -n²😉 2mn and (m² +n²😉
I suspect the answer is in there somewhere?
Are you impressed with me using the ² sign you told me about ?
I don't know how the silly faces got on there .they were supposed to be )
Managed to form a fairly horrendous quadratic, which gives 2.761 or 0.362. If they are correct, I guess you can have a steep or shallow ladder?
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25 Mar '21 20:27>
@blood-on-the-trackssaid Managed to form a fairly horrendous quadratic, which gives 3.761 or 1.362. If they are correct, I guess you can have a steep or shallow ladder?
There will be two solutions as you have indicated where the vertical distance above the box interchanges with the distance from the face of the box to base of the ladder. I'm asking for the distance above the box to the top of the ladder. You have given me a different distance, but if you subtract "1" from your solutions...hint hint.
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25 Mar '21 20:28>2 edits
@blood-on-the-trackssaid Managed to form a fairly horrendous quadratic, which gives 2.761 or 0.362. If they are correct, I guess you can have a steep or shallow ladder?
I see you've edited already before I've finished my reply!
Good job!
Blood On The Tracks has the correct answer(s)
I'll post my method in a little later this evening. If you wish to post your feel free.
y = [ ( 17^( ½ ) - 1 ) + ( ( 17^( ½ ) - 1 )² - 4 )^( ½ ) ]/2 ≈ 2.761 ft
and
y = [ ( 17^( ½ ) - 1 ) - ( ( 17^( ½ ) - 1 )² - 4 )^( ½ ) ]/2 ≈ 0.362 ft
And we are done! Please let me know if you you wish me to clarify anything in this ( its a bit challenging to read these equations in these forums - I've asked Russ for LaTeX capabilities...fingers crossed! )
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25 Mar '21 23:26>
@vendasaid I was never really interested in Trigonometry,but out of curiosity ,I looked on the net to see if it was possible to find angles in right angled triangles when given just one side measurement and it talked about something called Pythagorean triples with equations (m² -n²😉 2mn and (m² +n²😉
I suspect the answer is in there somewhere?
Are you impressed with me using the ² sign you told me about ?
I don't know how the silly faces got on there .they were supposed to be )
Are you impressed with me using the ² sign you told me about ?
good job venda! ( you were starting in the right direction )
"I don't know how the silly faces got on there .they were supposed to be"
Its a glitch. put a space following the ² before the closing parenthesis