A ladder and a 6 by 6 box ?

[phil_barker]

Hi,

A chap in the pub gave me a problem last night and I cannot solve it.

As I really need to get some work done, can anyone throw some light on the matter ?

 

A 6ft by 6ft box is placed against a wall, and a 20 ft ladder is placed such that it touches the wall, the ground and one corner of the box i.e (the ladder makes the hypothenuse). I need to calculate the opposite and the adjacent.

 

Regards,

 

Phil.

[YT2095]

is it a Rope ladder?

[ydoaPs]

WIll they be similar triangles?

[phil_barker]

Hi,

I think so. I reckon that the angle between the wall and the ladder is the same as the angle between the box and the ladder. And the angle between the ground and the ladder is the same as the angle between the 'top of the box' and the ladder.

 

Got as far as x^2 + y^2 + 24x + 24y = 328

 

but couldn't find another equation in terms of x and y in order to solve simoultaneously.

 

Regards,

 

Phil.

[ydoaPs]

Try drawing a picture.

[phil_barker]

I have attatched a diagram.

 

Cheers,

 

Phil.

Doc1.doc

[BhavinB]

Based on your diagram, X~17.84 ft and Y~9.04 ft (or the other way around...both answers are correct).

 

The key is to note the two similar triangles.

[psi20]

It'd be the other way around. X ~ 9.04 ft and Y ~ 17.84 ft. The top of the ladder would be 9.04 ft from the ground.

 

If you need to calculate the opposite and adjacent sides, you need to have an angle as a reference. But anyways, you got the numbers.

 

Or do you need to find the opposite and adjacent angles?

[phil_barker]

I'm still trying to work this out.

 

Based on similar triangles, I have found that:

 

 

X = 6Y/(Y-6)

The only other relationship I can see for X & Y is:

 

X^2 + Y^2 = 20^2

 

But I don't know how to solve these two.

 

Am I missing another equation here or do I need to solve the above two equations ?

 

Cheers,

 

Phil.

[BhavinB]

nope, thats it.

sub the first equation into the second and solve for Y.

 

To be fair...I used matlab to solve for the roots of the resulting fourth order polynomial.

[phil_barker]

Well, it could be the end of the road or the beginning of a new journey.

I've had a fiddle and subbed X = 6Y/(Y-6) into X^2 + Y^2 = 400

 

and out pops:

 

Y^4 - 12Y^3 - 328Y^2 + 4800Y - 14400 = 0

 

I can safely say that I don't know how to solve this, but if anyone has any links to suitable tutorials, it would be appreciated.

 

Thanks for your help,

 

Phil.

[psi20]

You can't solve this by hand (perhaps you could do it in your head). You can use Newton's Method, which requires some calculus and takes a long time to do. http://tutorial.math.lamar.edu/AllBrowsers/2413/NewtonsMethod.asp is a link to that.

BhavinB used matlab to solve it. I don't know how that program works.

I used a graphing calculator, graphed the function and pressed a few buttons to find when the graph becomes 0. The function above is in y. This can get confusing, because you need to put "Y^4 - 12Y^3 - 328Y^2 + 4800Y - 14400 = 0" as "X^4 - 12X^3 - 328X^2 + 4800X - 14400 = 0" into the calculator and solve for the zeros of the function. There are several roots but only one of the roots makes sense.

 

The equation should really be (Y^4 - 12Y^3 - 328Y^2 + 4800Y - 14400) / (y - 6)^2 = 0 . The roots are the same, but the graphs are different.

[alext87]

There are two similar triangles. Which means the angles must be the same.

 

so from the diagram drawn. tanθ=(x-1)/6 and tanθ=6/(y-6)

 

so 1/6xy - y - x + 6 = 6

x+y = 1/6xy also known is x2 +y2 = 400

 

therefore: 400 - 2xy = 1/36(xy)2

Solving this quadratic give xy = -36 + 12(SQRT109) = 89.28

 

sub in x= {-36 + 12(SQRT109)}/y

 

this gives the quadratic

y4 -400y2 +16992-864(SQRT109)=0

solving this gives:

y=19.47 so x=4.59 (3dp)

 

OR y=4.59 so x = 19.47 (2dp)

[The Thing]

A couple of methods I came up with all require solving of 4th degree polynomials, which I can't do by hand at all. But anyways, here are my methods:

 

Method 1:

Let x be the length of the smaller (bottom) triangle of the two similar triangles.

[math]

\frac{20-x}{6}=\frac{x}{\sqrt{x^2-6^2}}

[/math]

Simplifying this, including squaring both sides and removing the x from the denominator turns it into a 4th degree polynomial. I can't solve it.

 

Method 2:

Let x be the same as in Method 1.

[math]

sin^{-1}(\frac{6}{x})=cos^{-1}(\frac{6}{20-x})

[/math]

Turns out to be...something along the lines of a 4th degree polynomial.

 

Method 3:

Let X be the bottom left angle of both similar triangles.

Let x be the same as in Method 1 and 2.

[math]

sinX=\frac{6}{x}

[/math]

[math]

cosX=\frac{6}{20-x}

[/math]

[math]

\frac{6}{sinX}+\frac{6}{cosX}=x+20-x=20

[/math]

I dunno what this turns out to be. Haven't tried simplifying it yet but I'm not very optimistic about it turning out t be something other than a 4th degree polynomial again.