r = 5.6666666666666666666666666666667

Call the intersection of the two line segments A
Call the point where the line segment of length 5 intersects the circle B
Call the center of the circle C
Call the radius r

r = 3 + MYSTERYNUMBER

Now look at the triangle ABC

AB = 5
AC = MYSTERYNUMBER
BC = r = 3 + MYSTERYNUMBER

Go go gadget Pythagorean theorem:
BC² = AB² + AC²
(3 + M)² = 5² + M²
9 + 6M + M² = 25 + M²
6M = 16
M = 16/6 = 2.6666666666666666666667
...
r = M + 3
r = 2.666666666666666666667 + 3 = 5.666666666666666666667

Oh! Thank you very much. :D

Just something fun:
Find the tangent line of circle (x-a)²+(y-b)²=r² at point (c,d).

Draw a line from c, d (the point on the circle) to a, b (the center of the circle). The slope of this line would be perpendicular to the tangent line. Now you have the slope.

Slope of c,d-a,b = (b - d) / (a - c)
Perpendicular slope = (c - a) / (b - d)

That tangent line goes through c, d. Now you have a point.

(y - d) = slope * (x - c)
y - d = (c - a) / (b - d) * (x - c)

(y - d) * (b - d) = (c - a) * (x - c)

This thread makes me feel like an idiot...

This.

And nobody's even done any integration yet...

Oh, well how about ∫e^{x^2}dx?

I was thinking more along the lines of ∫x^xdx, but whatever.

Similarly difficult problems, no?

∫e^x^2dx
= ∫Σ((x^2k)/k!)
= Σ[(x^(2k+1))/((2k+1)k!)] + C

uh...I think...and k goes from 0 to ∞

oh and according to wolfram alpha, ∫x^xdx has no indefinite integral that can be represented with standard functions.

Even infinite series? That's intense.

Here's something I find a little funny. Sine, Cosine, and tangent are all use to describe angles. But when graphed, the look nothing like they would be involved with angles. Does anyone else find that funny?

I think they look like they're involved with angles... Then again, nothing math related really looks all that much like what it really is.

http://www.baselines.com/graphics/unit_circle.jpg

It's a Given. The two little dashy lines *tell* you that they're equal.

Actually, they only tell you that BC = CD.

I actually can't figure this one out. I mean, it's obvious from looking at it that they're equal, and probably you are going to use the Side-Angle-Side congruence property (er, postulate), but I can't figure out how to get that angle. There's probably a nice easy property about triangles inscribed in circles, but I can't remember it for the life of me.

Edit: Oh, got it. (looked up some properties)

In a circle, or congruent circles, congruent chords have congruent arcs. (and converse)
In a circle, or in congruent circles, two arcs are congruent if and only if the corresponding chords are congruent.

chord BC ≅ chord DC, so
arc BC ≅ arc DC, so
∠BAC ≅ ∠DAC

This was on my geometry test.
I found it quite difficult to get. Let's see if you can fare better.

Is that dot supposed to be the center?

z = 65 degrees and y = 25 if I am not mistaken.

And x is 50, of course.

Sorry, I flipped y and z around.

Solve this: