Thank you! So I would get 2 pi (r-t) if I subtract latter from former. But why do we need to do that? And where does dividing by s come in as theta = [2 pi (r-t)] / s?Using the arc-length formula, I would write:
[MATH]\overparen {XY}=(x+s)\theta=2\pi r[/MATH]
[MATH]\overparen {AB}=x\theta=2\pi t[/MATH]
Now, subtracting the latter from the former, what would we obtain?
I get theta = [2 pi (r - t)] / s. Wow Thank you so much!! Hmm I’m stil not sure about part b.. do I need to arrange it for x using similar triangles?You want to subtract not just the rightmost parts of the equations, but the two rightmost parts, so that you have:
[MATH](x+s)\theta-x\theta=2\pi r-2\pi t[/MATH]
What do you get from that, upon solving for \(\theta\)?
Thank you so, so much!! You’ve helped me so much.I would begin by stating:
[MATH]x=\frac{2\pi t}{\theta}[/MATH]
Now, substitute into that, the value you just found for \(\theta\), and simplify.
For the similar triangles, consider one right triangle with hypotenuse \(s+x\) and horizontal leg \(r\), and another with hypotenuse \(x\) and horizontal leg \(t\).