The swing traces a circular arc. 1.65m is a constant radius, whether it is straight down or off to the side.
Hi there.
This is a seemingly simple routine problem in my trigonometry book that I'm not sure about.
I've attached my paint recreation of the problem so hopefully you can see it.
In the book they explain that it is a swing that, when still is 0.35 meters above the ground, but when in "full swing" reaches h meters above the ground; how high is h?
My original answer didn't quite match up with the solution, but when I saw the answer I was somewhat confused with what they did.
If you look at the image on your left-hand side you can imagine how the space above 'h' could be called 'x' so that h + x corresponds to 1.65 + 0.35 on the other side (two parallel sides of 2). Now all you have to do is to figure out x so that you could calculate h = 2 - x
They did indeed do this, but then they reordered the geometry so that 'x' is part of the 72° triangle so that it could be calculated. My question is how did they draw a triangle, with the 1.65 side and the 72° angle, that at the same time included that x line.
In other words could; someone shed light on the transition between the geometry on the left to the right? Thanks.
Oh but of course. I must have stared myself blind on this one. Since they didn't mention it's circular arc I didn't assume it but of course the 'swing rope' is the same length at all times so it must be.
Thanks guys for the quick replies.