1. Originally Posted by Bernard.

I dun get how to do this. If you need anything cleared I can tell you what it is if my paint writing is that bad. The numbers are angle measures.
You can say since the center of the shape(?) U has 3 lines extending to the edges of the shape, the sides of the radii must be the same as well as the sides of all the triangles.
Imagine a hexagon with 6 triangles basically, all the sides are the same and consist of mainly the same angle measurements

2. just write fuck math on it

3. Originally Posted by TazWuzHereLowlz

You can say since the center of the shape(?) U has 3 lines extending to the edges of the shape, the sides of the radii must be the same as well as the sides of all the triangles.
This isn't a circle :\ and not necessarily 1/2 of a hexagon.

4. Angle-Side-Angle theorem or sth like that? Das some third grade shit can't remember.

5. Originally Posted by Bernard.

This isn't a circle :\ and not necessarily 1/2 of a hexagon.
Just wondering, is that a PART of the problem or is that the whole problem itself? It seems like an in-completed shape.

6. Originally Posted by TazWuzHereLowlz

Just wondering, is that a PART of the problem or is that the whole problem itself? It seems like an in-completed shape.
That's the whole problem itself.

7. Shit, I've failed geometry three times in row!

8. For HB: sin(22) = opp/hyp => HB/BU
For BL: sin(22) = opp/hyp => BL/BU

Therefore:
HB/BU = BL/BU
HB = BL

Disclaimer, not 100% sure because one of the triangles is over 90 degrees, but I assume it'll be fine.

9. ## The Following User Says Thank You to arunforce For This Useful Post:

Bernard. (02-11-2014)

10. Just look it up on Khan Academy and learn it. If you haven't learned this, you probably shouldn't be in your current grade of mathematics.
I miss this stuff. I can't even find my shit on Khan anymore. Awkward.

11. you need skype for a better explainations

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