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. 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.

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

4. 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.

5. 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.

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

7. 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.

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

[MPGH]Bernard. (02-11-2014)

9. 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.

10. you need skype for a better explainations

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