Chip

iTrader: (3)

I want a full algebra proof with your workings, not just your answer, and please no googling for it as there really isnt any point in that.

The answer is 2.4, you could easily find that out just by drawing and measuring, but I want to know how you got there with algebra


That just took me nearly 10 mins solid of algebra to work out, so anyone doing better than that is definately better at maths than me, lol

Im sure when I was at school that would have seemed a lot easier so I think im getting more stupid as I get older!

Ryan

surely it depends on the distance between the poles? so it means x is a variable, unless were taking those 2 triangle as isossceles or equalatrals?

StephenC

Id say you need one more variable, or as above to know if the angles of the two red triangles leave at the same angle top and bottom of the poles?

Chip

iTrader: (3)

No the distance makes no difference, it will be true for any distance other than zero (zero wouldnt make any sense anyway of course as x would be all values between 0 and the highest height)

But if you want a value for the distance between them, just use 1 as it will make your proof easier to work out.

Chip

iTrader: (3)

Its DEFINATELY possible to work out from just that as Ive just done it myself this morning.

StephenC

yeah, its 2.4 , will try write it up how I did it

CossieRich

iTrader: (1)

I have no idea what you are even try to work out. How embarassing for me

Chip

iTrader: (3)

Rich, the height of the intersection of the two lines, labelled X on this diagram.

Chip

iTrader: (3)

Still no one has PM'd me with a correct answer, come on folks, I know its friday and all that but its a great little problem to get stuck into I reckon

Ryan_3

its 2.4 mate.

pani_k

iTrader: (1)

At a guess since you can't solve it using trig it is to do with trapezium rule but if I had something like that I'd just scale it and solve it with a compass and protractor. Because I do so much uni level maths, I forget how to do things like this!

Chip

iTrader: (3)

Ryan, 0 marks for your workings there im afraid mate!

twinkle_2k86

i did it in 5 seconds, but cheated and used autocad.

still haven`t worked it out properly, can get to work out angles etc, even the ratio between them, just cant get the intersection point.

S1

Can I assume the two uprights are 90deg from the baseline?

FletchCossie

iTrader: (4)

The answer is 2.4, I worked it out by reading posts above, using my eyes.

Chip

iTrader: (3)

No good knowing things if you cant apply them.

And what if you had 100 to solve the distance for as part of a bridge you were building that had intersections like that, would you draw them all out individually rather than work out a formula?

Sorry, 0 marks again

Ryan_3

I wrote the working on a paper but the dog ate it, honest

Chip

iTrader: (3)

Yes, and so is X

You shouldnt need anything else other than pythagorus and basic algebra.

pani_k

iTrader: (1)

I've got a presentation to go and give now so when I return I'll sit down and try solve it! Not going to be defeated by simple trig and algebra when the maths I do makes this look like nursery level shit!!

Chip

iTrader: (3)

Lol, that was exactly my attitude, I was quite dismissive of how simple it was, but when I actually sat down to do it, it took me longer than I expected.

Big G

Ghostbusters?

Chip

iTrader: (3)

If the problem was "who can crack a shite joke" we would have several winners already

StephenC

using law of sines and basic trig its easy, just trying to write down a formula where you can put any value of x in for the baseline and it gives you that height, nearly got it, left my scientific calc at home though and windows calc in scientific mode is horrible

Chip

iTrader: (3)

Stephen, sounds like you are doing it a very different way to me, I didnt need a calculator at all, just windows notepad.
Im sure there will be more than one way to skin this particular cat though, so will be interesting to see how your working looks.

mk3matt

im at work... but i worked it out by knowing what the angle is within the triangle using trig... then used pithahorus ( i think that was his name lol) to get the height.

Chip

iTrader: (3)

I didnt use any angles at all, so again your method must be different to mine.

TimC

iTrader: (1)

Chip later on post your workings out on it.
Be interesting to see just where you started, as for me I need one more figure to even have a clue where to start. lol
I used to love this sort of stuff at school but my mind is drawing a blank since I aint done this sort of stuff in such a long time

alistairolsen

So you need to work out the ratios of x from each end, so you treat the bottom base as y, and the lengths from either end to x as y1 and y2. Using tangent you can resole them to (6y2/x) = ( 4y1/x) cancel x and you now know that the upright x is 3/5s of the way along the base from the left side.

if you now express it all in terms of y, so that the base is 2.5y and x is 1.5y from the left side you can then say for the smaller triangle (side 4) that tan (theta) = 4/2.5y and also x/1.5y which once you cancel the theta terms gives x=2.4.

daviddunlop83

iTrader: (7)

Long time ago since i did any of that stuff, tho doing it on cad is def quicker and you can change the parameters very quickly ifyou wished

StephenC

Well, I got the answer by trial and error using various values for the baseline,
these are my workings



used this link
http://ostermiller.org/calc/triangle.html
which is great for any triangle calcs

Not gonna be able to come up with just one equation, although it should be possible.

think I may have looked too deep to solve it if you did it with basic pythag and algebro. Always was my problem at college though, looking beyond the obvious.

Chip

iTrader: (3)

Intersting Ali, totally different way to how I did it.

I derived a formula that you can put the values for the left and right height in which comes out with the result.
The formula is

L * R / ( L + R )

So in this case
24/10 = 2.4

Which is a more simple to use equation in future, but no more valid than how you did it of course



The way I got to that formula was:

distance between the poles is A

distance from X to the right pole is proportinal to the left and right poles heights, so X to the left pole is A * R/(L+R)

hypotenuse for the triangle of height LEFT is
the square root of (L*L + a*a)

hypotenuse of the smaller triangle of height X is a fraction of R/(L+R) of that

hypotenuse of the smaller triangle
R/(L+R) * root of (L*L + A*A)

base of the smaller triangle is R/(L+R) * A

square of hypotenuse - square of base is square of X

so square of X =
(R/(L+R) * root of (L*L + A*A) * R/(L+R) * root of (L*L + A*A)) - (R/(L+R) * R/(L+R))

So X squared + (R/(L+R) * R/(L+R)) = (R/(L+R) * R/(L+R)) * root of (L*L + A*A) * root og (L*L + A*A)

Substitute 1 for A and simplify

x squared + (R/(L+R) * R/(L+R)) = (R/(L+R) * R/(L+R)) * (L*L + 1)

simplify further

x squared + (R/(L+R) * (R/(L+R)) = (R/(L+R) * R/(L+R)) * (L*L)) + (R/(L+R) * R/(L+R))

subtract (R/(L+R) * R/(L+R)) from both sides

X squared = (R/(L+R) * R/(L+R)) * (L*L)

root both sides

x = R/(L+R) * L
X = R*L/(L+R)

ballin

iTrader: (1)

fuck this, im going to the beach

DanW@FastFord

iTrader: (1)

You did that shit for fun???

I'd rather rub my knob with a cheese grater than do algebra!

Chip

iTrader: (3)

Yes, I enjoy problem solving, mathematical or otherwise.

And to be fair its not like maths has been useless to me in life either, I currently make a fairly good living writing finance software for example, so its a handy thing to enjoy.

StephenC

whats the next one then chip???

Chip

iTrader: (3)

I didnt make it up, GarethT sent me it.

DanW@FastFord

iTrader: (1)

Problem solving - yes, Maths - No! I'm just grateful the only Maths I have to do is to work out how many pages my mag has!

Chip

iTrader: (3)

Its always an even number so ignore all the odd numbers when you are working it out, that should save you some time

Psycho Warren

iTrader: (5)

what a waste of 15 minutes of my life Chip. Thanks

I did it same way as you to avoid having to use sines, cosines and tangents which just complicate things needlessly.

Chip

iTrader: (3)

Also it leads to a neater more useable final equation done "our" way, although its not like thats ever going to be any use to me anyway