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Math mens come
 | 
Brazil hrp__ 
imgur.com/BvYHB6z Is there a way to find h ? I mean, yes I could split the big triangle in two right angle triangles and then solve for h/2 that would be equal to sine (45/2) * 2 if I am not wrong. But I gotta do it without using trigonometry. So I wonder if there is a way do it. If you know how, pls help me mens. btw, A is a point in the coordenates (0,0) and B is in (1,0). The lenght of the arc B B' is 0.785 if im not wrong.
2020-03-28 13:09
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#1
SolEk | 
Europe Adenauer 
I wish i was that smart !
2020-03-28 13:10
#2
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Brazil hrp__ 
+1
2020-03-28 13:10
#41
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Estonia MorsAlbum 
yes? just fucking count??
2020-03-28 13:55
#42
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Brazil hrp__ 
wdym
2020-03-28 13:55
#43
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Estonia MorsAlbum 
0,0 to 0,1 is 10 squares aka 0.1s, just count and use logic lmao edit: the first 'h' on the left has a slope of 1, the 2nd 'h' which you're trying to find has a slope of -2 aka h is 0.5 because its exactly half of the left side which is (1)
2020-03-28 13:57
#44
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Brazil hrp__ 
I want h not AB men))
2020-03-28 13:57
#45
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Estonia MorsAlbum 
read
2020-03-28 13:57
#47
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Estonia MorsAlbum 
no its not exactly half nvm but its along those lines which i cant be fucked to do since i left school 2 years ago lmao. basically just find the slopes using logic and make an equation. the equation is some shit like point A (x,y) * 2(m aka slope) or some shit like that, do it for the left triangle and then equate it to the right triangle and solve for the right triangle's "h"
2020-03-28 13:59
#3
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Sweden friendly_swede 
I mean, Pythagorean theorem (and sin, tan, cos etc.) is the only way I can see it be done. I haven't had coffee today though so I might be temporarily retarded.
2020-03-28 13:14
#5
ScreaM | 
Brazil HLTVolt 
OP said without Trigo
2020-03-28 13:16
I know
2020-03-28 13:22
#11
ScreaM | 
Brazil HLTVolt 
I did had coffee and yet I miss read your comment, looks like I'm legit retard
2020-03-28 13:22
Haha, no worries mate. We've all been there :D
2020-03-28 13:24
#7
 | 
Brazil hrp__ 
I don't see how I can solve it using the Pythagorean theorem though :( I mean if I split the triangle ABB' in two, I'd get a triangle with two unknown values: h/2 and the other side's lenght
2020-03-28 13:17
NVM, didn't see that the side j doesn't line up.
2020-03-28 13:23
xdd
2020-03-28 14:43
#4
ScreaM | 
Brazil HLTVolt 
Just count the squares? 3 squares horizontal 7 squares vertical h = sqrt(0,3² + 0,7²)
2020-03-28 13:15
#8
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Brazil hrp__ 
well I could do it but its a bit off the actual value
2020-03-28 13:19
What's the entire question? The triangle is inscribed in a circle. You can find the angles using the vertical line as a reference to create 2 triangles, so you know you have two 45 degree angles and a 90 degree angle. You can also just use the grid in the back to know how many units long the line is. If it doesn't give you any units or any measurements to work with though you can't find it's length because it could just be dilated
2020-03-28 13:16
#14
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Brazil hrp__ 
Well its an isosceles triangle with side lenght 1 with BAB' making an angle of 45 degrees and I should solve for the lenght of its other side. Using the grid is a good idea but that would be a gross approximation tbh
2020-03-28 13:24
#9
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Libya s1v9mple 
Fuck geometry
2020-03-28 13:22
#23
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Bulgaria ohmeingutness 
1st time i agree with you on smthing here
2020-03-28 13:27
#28
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Libya s1v9mple 
Never had any issues with any algebra/calculus or trig, but fucking geometry. I hate it so fucking much.
2020-03-28 13:32
#31
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Bulgaria ohmeingutness 
trig and geo for me, worstest
2020-03-28 13:33
just use a ruler menss
2020-03-28 13:24
#17
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Brazil hrp__ 
my ruler is too big))
2020-03-28 13:25
Use trigonometry. And sin(A/2) = sqrt((1 - cos(A))/2)
2020-03-28 13:25
#19
 | 
Brazil hrp__ 
no men)) there is a way of doing this without trigonometry (i guess)
2020-03-28 13:26
Just so I'm understanding correctly, you want to calculate the chord length without using trigonometry?
2020-03-28 13:25
#20
 | 
Brazil hrp__ 
yes, thats it.
2020-03-28 13:27
Look at #26, he summed it up pretty well for you.
2020-03-28 13:33
find cords and then count the length ez
2020-03-28 13:27
#22
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Brazil hrp__ 
how ? help pls
2020-03-28 13:27
you can easily find cords of B and B' (triangle has equal sides and piphagorian theorem) then there is a formula for length between two points: r = (x1 - x2)^2 + (y1 - y2)^2
2020-03-28 13:33
#24
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Romania ViolentJ 
Cos theoreme: (BB`)^2=(AB)^2+(AB`)^2-2(AB)(AB`) x cos A
2020-03-28 13:28
#27
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Brazil hrp__ 
without using cos sin or tg mens))
2020-03-28 13:31
#50
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Romania ViolentJ 
well shit mens, gl with that
2020-03-28 14:03
#25
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Bulgaria ohmeingutness 
idk man i failed my geometry and trigonometry shit, i now understand a bit but thats some new high level shit
2020-03-28 13:28
#26
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Libya s1v9mple 
Anyway, the triangle made by A, B1 and that unspecified point is an isosceles triangle, since y = 90 and A = 45, so B1 is 180-90-45 = 45. Just remember this for any time you spot a right triangle with one angle being 46. So the short, equal sides are of some length, x, and by the Pythagorean theorem, the hypotenuse, which is 1, gives the equation x^2 + x^2 = 1^2, in other words, x = 0.5sqrt(2). AB1 = AB, since A is the middlepoint of a circle and B1 and B are both on it, so j = x = 0.5sqrt(2) and that short length on AB is 1- 0.5sqrt(2). Now use the Pythagorean theorem again. h^2 = (0.5sqrt(2))^3 + (1-0.5sqrt(2))^2 and you can simplify that.
2020-03-28 13:30
+1, GJ, you're the one person who is able to read instructions and not use trig for the answer
2020-03-28 13:34
#33
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Brazil hrp__ 
thats cool but how do you assume j = 1 - the segment going of unspecified point to B
2020-03-28 13:36
#34
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Libya s1v9mple 
J is the same as the large segment of AB (so x), since it's an isosceles triangle. I'm pretty sure I didn't say it was 1- something.
2020-03-28 13:40
#38
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Brazil hrp__ 
oh you right ! thank you. Anyway, I mean the distance of the unspecified point to B is equal to 1 - j. Do you think there is a way to generalize it ? I mean, if the angle is not 45 in the triangle B'BA it is impossible to solve ?
2020-03-28 13:51
#53
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Libya s1v9mple 
If it was a 30-60-90 trisngle, then you could actually figure out the sizes of the sides, you can look it up. 45-45-90 and 30-60-90 triangles are special cases. In any other case, you'd need trig.
2020-03-28 14:06
#56
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Brazil hrp__ 
Cool. Thank you. Anyway I guess there is a way of doing it for every case without using trig !
2020-03-28 14:08
im retarded
2020-03-28 13:43
just use this formula men chordlength = 2* radius*sin (A/2) i got 0.765 units
2020-03-28 13:43
#37
ScreaM | 
Brazil HLTVolt 
You can develop the line like this and use Talhes to find the rest without trigo i.imgur.com/eHSgFvl.png
2020-03-28 13:43
#40
 | 
Brazil hrp__ 
thank you ! but how do you know B' to the extended point is equal sqrt(2) ?
2020-03-28 13:53
#46
ScreaM | 
Brazil HLTVolt 
Because of the angle, it's 45°, if you go 3 squares to the right from your B', then 3 squares up, then count the squares from where you are all the way to the B angle, you get 1 Pythagoras then : sqrt(1+1) = sqrt(2)
2020-03-28 13:59
#49
ScreaM | 
Brazil HLTVolt 
using Thales the line below is 0,70710678118654752440084436210485 for the first triangle and then 0,29289321881345247559915563789515‬ for the second triangle, if add them it's = 1 to prove that it is correct we can use pythagoras to find the "j" line now by using : 1 = sqrt( j² + 0,7...²) 1² = j² + 0,7...² j = sqrt (1² - 0,7...²) j = 0,70710678118654752440084436210485 j is the same as the line below that we found you can now search for "h" using pythagoras once again h = sqrt (0,7...² + 0,2...²) h = 0,7653668647301795434569199680608‬
2020-03-28 14:02
#51
 | 
Brazil hrp__ 
you're right. Thank you. Do you think there is a way to generalize it ? I mean find h for every possible angle (0, to close to 90) of the ABB' triangle ?
2020-03-28 14:05
#59
ScreaM | 
Brazil HLTVolt 
Without trigo???
2020-03-28 14:11
#61
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Brazil hrp__ 
yes, I mean using trigonometry is easy
2020-03-28 14:12
you can use sqrt(sin(x)²+(1-sin(x))²)
2020-03-28 14:19
#65
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Brazil hrp__ 
yeah, but without using trigonometry seems to be more fun))
2020-03-28 14:15
I dont know how to make a general formula without trig((
2020-03-28 14:15
#70
ScreaM | 
Brazil HLTVolt 
Same here, I don't know what the fuck his teacher is doing but wtf is this
2020-03-28 14:20
this is correct but you dont even need to know about the principles because we know that j is sqrt(2) and the base between A and (0,0) is sqrt(2) just using pythagorean theorem you get sqrt(sqrt(2)²)+(1-(sqrt(2))²) = 0.765...
2020-03-28 14:20
I am not sure if I can explain with words but the distance AB is 1, so the height also must be one because we know that the other angle must be 45 degrees, therefore it is sqrt(1^2 + 1^2) = sqrt(2)
2020-03-28 14:06
#57
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Brazil hrp__ 
omg im dumb. I didn't notice the other side also was 1 because of the angle.. thank you !
2020-03-28 14:09
no worries hrp boss
2020-03-28 14:11
#62
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Brazil hrp__ 
I am glad i didn't choose maths for my life !
2020-03-28 14:13
#39
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Turkey Cormack 
Bro how can ı open fucking imgur sites
2020-03-28 13:53
#52
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Brazil hrp__ 
idk maybe turkey blocked it ?
2020-03-28 14:05
rip Harvard for me
2020-03-28 14:01
#55
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Europe obvious 
Just help yourself with the squares, use pythagorean theorem. I'm sure it would be pretty easy. Do you know the length of a square or something? If so then you can easily solve it. If 1 is the length of |AB| then then one square is 0,1 and a square inside that square would be 0,02. That's all the info you need to solve it.
2020-03-28 14:14
thats true but j is not exactly on the line and neither is the top point so it will be slightly off, what you are saying would give sqrt(0,7²+0,3²) instead of sqrt(sqrt(2)²+(1-sqrt(2))²)
2020-03-28 14:18
#72
 | 
Europe obvious 
True, didn't notice that it's slightly off.
2020-03-28 14:21
#63
magixx | 
Russia avvakum 
i.imgur.com/qujiU38.png r8 my handwriting, i think its a solid 9/8
2020-03-28 14:13
#83
magixx | 
Russia avvakum 
2020-03-28 14:39
So easy that I can't stop laughing at retards on hltv. Have you heard about something know as similarity and pythagoras theorem? Let the foot of perpendicular be P Then triangle AB'P is similar to triangle B'BP So j^2=x*(1-x) where x is length of AP. Now find value of j^2 from 2 right triangles and this value. Equate all 3, factorise them and done.
2020-03-28 14:24
#71
ScreaM | 
Brazil HLTVolt 
I laugh at disabled kids who can't read
2020-03-28 14:20
first answer is without trigonometry because I'm not blind. I would never produce that solution if there were no restrictions. Second one is obvious solution and straight forward. Just to show you beauty of maths.
2020-03-28 14:23
#78
ScreaM | 
Brazil HLTVolt 
You make me wanna commit suicide
2020-03-28 14:28
wasn't my intention sorry maybe I didn't understand you
2020-03-28 14:29
#80
ScreaM | 
Brazil HLTVolt 
was indeed your intention, omw to never see this kind of posts again, bye!
2020-03-28 14:29
Happy?
2020-03-28 14:24
#75
 | 
Brazil hrp__ 
well i forgot to mention I was looking for a generalization. How do I solve it for other angles ?
2020-03-28 14:24
well I edited my answer Now you will see only generalised answer.
2020-03-28 14:25
#84
 | 
India baba_dhanraj 
ggwp
2020-03-28 14:45
is point a the center of the circle slice?
2020-03-28 14:19
#77
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India baba_dhanraj 
pythagoras in left triangle a^2 + a^2 = 1 find a j = a; in right triangle base = 1-a; now h^2 = j^2 + (1-a)^2; ez
2020-03-28 14:27
what??? wow your brain is bigger than striders
2020-03-28 14:32
#82
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Lithuania Sedge 
if my calculations are correct the answer is 7.
2020-03-28 14:37
0.764 is the answer
2020-03-28 15:10
#87
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Lithuania Sedge 
yes so basically i am right, maths is all about approximations.
2020-03-28 16:12
nice approximation
2020-03-28 17:03
#88
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Canada ProvexPyker 
Use Pythagorean to find bigger length of AB, once you know that then the short side of AB is just 1 - bigger side. Now you have the small side of AB and you can use pyth theorem to find h
2020-03-28 16:39
trash bump
2020-04-02 23:24
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2020-04-04 17:06
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