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Dragon Ball Super Discussion Thread 100

MeiouHades said:
Anyway, let’s try this again. Time for some math.
XKkqq.jpg

3 congruent squares. Find the sum of angles A+B+C. Show your work.
Click to expand...
Try this instead
 
I feel like things would be less negative if DBS just had its head pulled out of its arse. We're still stuck on SH after over a year and the anime still isn't back in any sense, I've even heard the SDBH anime may be ending soon. We have an ongoing series where absolutely nothing is happening. Like damn, I just want to be out of this filler hell and back into a new arc with new techniques, new feats, new scaling, new anything.
 
Vietthai96 said:
let be real here, aside from the Low 1-C thread which i'm still participating, i care more about gathering scans and references for the verse. when I active on this wiki, like almost all DB profiles have no scan, can we focus on update the quality of our verse profiles???
Click to expand...
almost like that's what I've been suggesting for a while now
 
In other thought, C looks like 45°, B should be lowered, 30 or 25°, A 15°, so calculating the sum will be 90°
 
ImmortalDread said:
In other thought, C looks like 45°, B should be lowered, 30 or 25°, A 15°, so calculating the sum will be 90°
Click to expand...
C should be 45 for cutting the square in two perfect triangles.
For B and A just slap 1 on each side of the square and you have two different triangles and use pythagorium to find all sides. I don't know about geometry but I would think knowing about the sides would give the angle. B triangle is basically 1^2+2^2= root5 and A triangle is 1^2 + 3^2=root 10

No idea if there exists a better way to do it. Thinking about it.
 
They're all right triangles. C is trivially 45 because it bisects a square diagonally, but for the other triangles you need a big of trigonometry. The squares are the same distance so you know the ratios of the sides, so if you treat the sides of the square as "1" then you can get the ratios of the non-hypotenuse sides very easily and use an inverse tangent function to get the angle in question, but the numbers aren't nice pretty numbers in that case, but A and B add up to 45 so the final result is easier.

A = 18.4349488 degrees
B = 26.5650512 degrees
C = 45 degrees

A + B + C = 90 degrees.

There might be an easier way to do it, but all I remember from trig is Pythagorean theorem and SOH-CAH-TOA so that's the best I can do lol.
 
Deagonx said:
They're all right triangles. C is trivially 45 because it bisects a square diagonally, but for the other triangles you need a big of trigonometry. The squares are the same distance so you know the ratios of the sides, so if you treat the sides of the square as "1" then you can get the ratios of the non-hypotenuse sides very easily and use an inverse tangent function to get the angle in question, but the numbers aren't nice pretty numbers in that case, but A and B add up to 45 so the final result is easier.

A = 18.4349488 degrees
B = 26.5650512 degrees
C = 45 degrees

A + B + C = 90 degrees.

There might be an easier way to do it, but all I remember from trig is Pythagorean theorem and SOH-CAH-TOA so that's the best I can do lol.
Click to expand...
I was somewhat correct?
 
You were almost there. You calculated the hypotenuse correctly but you can actually solve it without the hypotenuse.
 
Deagonx said:
They're all right triangles. C is trivially 45 because it bisects a square diagonally, but for the other triangles you need a big of trigonometry. The squares are the same distance so you know the ratios of the sides, so if you treat the sides of the square as "1" then you can get the ratios of the non-hypotenuse sides very easily and use an inverse tangent function to get the angle in question, but the numbers aren't nice pretty numbers in that case, but A and B add up to 45 so the final result is easier.

A = 18.4349488 degrees
B = 26.5650512 degrees
C = 45 degrees

A + B + C = 90 degrees.

There might be an easier way to do it, but all I remember from trig is Pythagorean theorem and SOH-CAH-TOA so that's the best I can do lol.
Click to expand...
This is correct. However there is a way to do it without any trig if you don’t remember much from it, like you said.
 
Deagonx said:
They're all right triangles. C is trivially 45 because it bisects a square diagonally, but for the other triangles you need a big of trigonometry. The squares are the same distance so you know the ratios of the sides, so if you treat the sides of the square as "1" then you can get the ratios of the non-hypotenuse sides very easily and use an inverse tangent function to get the angle in question, but the numbers aren't nice pretty numbers in that case, but A and B add up to 45 so the final result is easier.

A = 18.4349488 degrees
B = 26.5650512 degrees
C = 45 degrees

A + B + C = 90 degrees.

There might be an easier way to do it, but all I remember from trig is Pythagorean theorem and SOH-CAH-TOA so that's the best I can do lol.
Click to expand...
My method was way easier and I got 90° smh
 
There is no objective “correct” method except measuring it. There are three unknown values, with a sum also being unknown to us. The only way to solve that is to measure it in fact, since the problem is visual.

I simply guessed it because I solved those type of angles, 10000 times to be able to guess their angle approximately correct.
 
ImmortalDread said:
There is no objective “correct” method except measuring it. There are three unknown values, with a sum also being unknown to us. The only way to solve that is to measure it in fact.

I simply guessed it because I solved those type of angles, 10000 times to be able to guess their angle approximately correct.
Click to expand...
Given a multiple choice I would also pick 90 but I would prefer to have actual numbers if time is not a problem.
 
I'm sure there's a way to intuit that A and B add up to 45, but I don't know how to do so.
 
Deagonx said:
Tbh I don't follow the logic here.
Click to expand...
You construct 6 additional congruent squares as shown, then you construct the right triangles with angle X and Y vertically to notice that they’re part of an isosceles right triangle, meaning they add up to 45. That means X+Y+Z=45+45=90
 
pineappleman said:
Please don't make a blog just yet, I see some room from improvement (stuff like SFPSSJ4 Goku scaling to Omega, and not Syn/EoGT Goku scaling above Omegaa in base) and i'm planning on doing a full rewatch of GT to get everything more ironed out/see if anything can be added
Click to expand...
Wait I am confused you would scale SFPSSJ4 to Omega or to Syn?
 
If this problem stumped you, don’t feel too bad. It also stumped by college mates, they’re mathematics majors.
 
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