World's Hardest Easy Geometry Problem
- Thread starter Crystal Cuckoo
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and we can assume failing said subject as well!
Common sense and simple maths is all that's needed.
Sum of angles in a triangle = 180.
From that you can resolve most of the missing angles. Once you've done that you'll find that the angle below X equals 30 and that the sum of X plus the angle above equals 150. You'll also have found that all the angles around X end with 0. The drawing is to scale and it is obvious that X is smaller than the angle below it so X must be less than 30. So from this we have
a). X must end with 0
b). X must be less than 30
c). The sum of the angles must resolve to equal 180.
From this you can determine that the value of X must be either 20 or 10. All you need to do is check that the once X has been substituted with either 20 or 10 that the sum of the angles around it all add up to 180. The result being that X = 20.
Or you could waste time writing out Algebra and drawing extra lines to make more triangles with yet more angles. Whatever works for you.
Are AE and BD straight? I "could" assume they are but you haven't specified. Just because they look straight doesn't mean they are.
Check the scale diagram... It's assumed that those two lines are straight...Hunter_Silver said:
Are AE and BD straight? I "could" assume they are but you haven't specified. Just because they look straight doesn't mean they are.
Yup, I got 20 degrees to. However I don't know how to put a pic into my thread (yeah, yeah, serious case of n00b) but it's an easy question with some basic math. Extra lines and algebra? Didn't seem like you needed to do all that but whatever works. I just did lots of adding and subtracting.
Hmm I'm ussually good at this shit, I just couldn't remember how to solve the 30/110, 20/130 angles in the CED triangles.
I knew AEC, angle E = 150, and CDB, with D obviously being 140. (20+20+140 = 180)
But solving EDB or AED don't work right. Of course I've totally forgotten most of the opposite angles rules and such things.
Ironicallly, It should be easy, as you can solve the 50/50/130/130 straight away, and X and that other angle = 50, and since its easy to figure out that one angle = 30. I believe theres a rule out there that should point that sine that one angle = 50, those two angles combined = 50, and since we know one = 30, we can deduce that the opposite = 20.
I knew AEC, angle E = 150, and CDB, with D obviously being 140. (20+20+140 = 180)
But solving EDB or AED don't work right. Of course I've totally forgotten most of the opposite angles rules and such things.
Ironicallly, It should be easy, as you can solve the 50/50/130/130 straight away, and X and that other angle = 50, and since its easy to figure out that one angle = 30. I believe theres a rule out there that should point that sine that one angle = 50, those two angles combined = 50, and since we know one = 30, we can deduce that the opposite = 20.
the answer is 40, at least i think it is. I didnt look at any of the hints and did it in my head, so i could be wrong. but i think i'm right, when i took geometry, i just missed one SOL question on it (but dont call me a nerd). either way, i cant believe i took time to do this during spring break.