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August 26th, 2015, 12:17 AM   #1
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Finding the minimum length of l using trig functions

The Problem given was that there's a house of unimportant height and wall 10ft tall 8ft from the house. Now, a ladder is placed such that one end of the ladder rests on a point on the wall of the house and the base rests outside the wall.

A figure is given (see attachment).
The left line represents the house.

If there are points that aren't clear, feel free to ask.

Any explanation is greatly appreciated.
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Last edited by skipjack; August 26th, 2015 at 01:14 PM.
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August 26th, 2015, 02:34 AM   #2
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What are you asked to do or to explain?
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August 26th, 2015, 06:41 AM   #3
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It's in the title.

We have to find the minimum length of the ladder using trigonometric functions.
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August 26th, 2015, 08:54 AM   #4
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Quote:
Originally Posted by Volle View Post
The Problem given was that there's a house of unimportant height and wall 10ft tall 8ft from the house. Now, a ladder is placed such that one end of the ladder rests on a point on the wall of the house and the base rests outside the wall.

A figure is given (see attachment).
The left line represents the house.

If there are point that aren't clear, feel free to ask.

Any explanation is greatly appreciated.
Approximately 36.1987 ft. On the other hand, your diagram shows the ladder going to the top of the house. If that is a requirement then the height of the house goes from unimportant to very important.
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August 26th, 2015, 10:20 AM   #5
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Are you sure?
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August 26th, 2015, 10:53 AM   #6
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Are you sure?
How about 25.4033? I think I made a rookie mistake. My angle was in picomils rather than radians.
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August 26th, 2015, 12:15 PM   #7
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Using similar triangles and a graphing calculator, I get $l$ $\approx$ 25.4033 feet.

Volle, did you have a particular approach in mind?
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August 26th, 2015, 01:06 PM   #8
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Let θ be the acute angle of elevation of the ladder, then its length l = (8sec θ + 10csc θ) ft.

To locate the minimum, we solve dl/dθ = 0.

We need 8sec θ tan θ - 10cot θ csc θ = 0, so tan³ θ = 5/4.

Plugging the appropriate value of θ into the equation for l gives l = 25.4 ft after rounding to 3 significant figures. An exact expression can be found for this, but is inelegant.
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August 26th, 2015, 03:03 PM   #9
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Originally Posted by mrtwhs View Post
Approximately 36.1987 ft. On the other hand, your diagram shows the ladder going to the top of the house. If that is a requirement then the height of the house goes from unimportant to very important.
I guess that's my bad in the diagram. I wanted to mean that one end of the ladder could be at any point on the house, not necessarily on top of it.

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Originally Posted by greg1313 View Post
Using similar triangles and a graphing calculator, I get $l$ $\approx$ 25.4033 feet.

Volle, did you have a particular approach in mind?
I did. The only idea that came to me was using similar triangles but then it was immediately quashed because although I used trig functions (sin, cos, and tan), I couldn't arrive to any good answer.



Quote:
Originally Posted by skipjack View Post
Let θ be the acute angle of elevation of the ladder, then its length l = (8sec θ + 10csc θ) ft.

To locate the minimum, we solve dl/dθ = 0.

We need 8sec θ tan θ - 10cot θ csc θ = 0, so tan³ θ = 5/4.

Plugging the appropriate value of θ into the equation for l gives l = 25.4 ft after rounding to 3 significant figures. An exact expression can be found for this, but is inelegant.
This is exactly the solution I'm looking for. Thank you very much. This is very interesting to me. I've never thought of using the reciprocal identities.
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