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.

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.

I have attatched a diagram. Cheers, Phil. Doc1.doc

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.

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.