Frustum of a cone question

[Ocean]

I’m having a lot of trouble with this question.. i’ve found arc length for part a but I have no idea how to show in radians theta = [2 pi (r - t)] / s as well as part bi and bii.

 

[MarkFL]

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?

 

[Ocean]

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?

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?

 

[MarkFL]

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 \(\theta\)?

 

[Ocean]

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 \(\theta\)?

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?

 

[MarkFL]

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 \(t\).

 

[Ocean]

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 \(t\).

Thank you so, so much!! You’ve helped me so much.