Tangent & Radius are Perpendicular (Proof by Contradiction)

  Рет қаралды 145,739

Eddie Woo

Eddie Woo

Күн бұрын

Пікірлер: 124
@Mech.Masters
@Mech.Masters 6 жыл бұрын
I can feel the love you have for learning and teaching maths. Keep it up sir
@angelopinheiro3966
@angelopinheiro3966 5 жыл бұрын
Just look how straight that Ox line is with no ruler
@mikedebell2242
@mikedebell2242 5 жыл бұрын
Man! He drew a STRAIT line from O to x intersecting the circle at b without a ruler! That's talent!
@goldwinger5434
@goldwinger5434 4 жыл бұрын
Actually, he drew a straight line.
@JTtheking134
@JTtheking134 3 жыл бұрын
tell me if you can make a line O to x without intersecting the circle though...
@carlibouros
@carlibouros 2 жыл бұрын
@@JTtheking134 hum, the point is that the line is straight, no one cares about the fact that it is intersecting the circle 😆
@warplanner8852
@warplanner8852 2 жыл бұрын
@@goldwinger5434 and Mike is in _dire_ straits.
@joshuawebb6444
@joshuawebb6444 2 жыл бұрын
@@warplanner8852 One could say he is a tangent 'Brother in Arms'...
@tristanmoller9498
@tristanmoller9498 6 жыл бұрын
Finally a math teacher that doesn’t start a lesson by saying: “Ok, this is going to be boring...” No shit math ain’t boring
@azmanmatamin9020
@azmanmatamin9020 5 жыл бұрын
Ikr they need passion in it
@ngrobert5054
@ngrobert5054 4 жыл бұрын
true not boring
@goldwinger5434
@goldwinger5434 4 жыл бұрын
I got my degree in math about forty years ago and I love this channel. He goes over a lot of stuff that I took for granted for decades and I sit here and go "Wow! I never knew that." For example, I've known about sines and cosines since 10th grade trig BUT no one ever explained that cosine is shorthand for the complement of sine. Also, I now see the relationship between sine and sinuous (I'm fascinated by words and their origins.).
@goldwinger5434
@goldwinger5434 4 жыл бұрын
@Rusty Never really gave it any thought but origin of words was never covered.
@InTheBeginningTheUniverseWas
@InTheBeginningTheUniverseWas 5 жыл бұрын
I like that if BX is negative, you can imagine X is inside the circle. Therefore the line PQ cuts the circle in two places, so it isn't a tangent... which is a contradiction because we defined it as such in the set up.
@gyedublay
@gyedublay 4 жыл бұрын
Genius
@scotthix2926
@scotthix2926 4 жыл бұрын
That is what I was thinking
@reubenmanzo2054
@reubenmanzo2054 3 жыл бұрын
The fact still stands that BX is distance, not displacement. Since it is a scalar, not a vector, it has no direction. Without direction, it can't possibly have a negative magnitude.
@briankelly5828
@briankelly5828 6 жыл бұрын
Outstanding teacher.
@peterjohnson4854
@peterjohnson4854 6 жыл бұрын
You have a talent for teaching. Keep up the good job.
@nicholasc5740
@nicholasc5740 3 жыл бұрын
He answered every question I had in mind except how he drew such a perfect circle
@qayyimfarts627
@qayyimfarts627 5 жыл бұрын
5:37 just pointing it out
@santhoshbejugama7318
@santhoshbejugama7318 3 жыл бұрын
First time, I am interested in Mathematics . Super class
@andrewfarquhar9083
@andrewfarquhar9083 2 жыл бұрын
I wish I had this teacher so captivating and willing to explain.
@easportsforever
@easportsforever 8 жыл бұрын
Brilliantly explained!
@smithmeister
@smithmeister 4 жыл бұрын
I wish I had this guy teaching me maths in school I swear to god!
@bruce33331
@bruce33331 3 жыл бұрын
lovely explanation ...... this man is a maestro
@Name-ps9fx
@Name-ps9fx 2 жыл бұрын
I had a friend who was very interested in art, and one of the things he did almost every day was practice drawing circles, straight lines, squares etc. Not for hours, mind you, but just 5-10 minutes each day. He obviously got very good at drawing these basic shapes without a straight edge, ruler, or compass. Point is, if you want to have a skill, you have to practice it.
@murattanyel1029
@murattanyel1029 4 жыл бұрын
Correction: In proof by contradiction, you don't start with a false premise. You start with a premise you are testing.
@HansWeberHimself
@HansWeberHimself 3 жыл бұрын
No you don’t!
@MadaraUchihaSecondRikudo
@MadaraUchihaSecondRikudo 3 жыл бұрын
@@HansWeberHimself Let's assume that you do
@HansWeberHimself
@HansWeberHimself 3 жыл бұрын
@@MadaraUchihaSecondRikudo Funny, but no. 😉
@TheFirstNamelessOne
@TheFirstNamelessOne 2 жыл бұрын
@@HansWeberHimself If you knew it was false, what's the point?
@HansWeberHimself
@HansWeberHimself 2 жыл бұрын
@@TheFirstNamelessOne It’s not false or true by absolute measures, it’s false in relation to the hypothesis. Example: Probably a bad one as I commented on his a while ago: hypotheses is that my car is red. Proof by contradiction starts with the assumption that my car has no color. We then show that my car must have color and further discover that that color could be red. That’s usually good enough to make the needed point. My car has a color. Red is an option. Then you just need a picture.
@khotsok2491
@khotsok2491 4 жыл бұрын
Kid coughs in 2015: carries on* Kid coughs in 2020: everyone, get out get out!
@mariazuza2593
@mariazuza2593 4 жыл бұрын
omg, he can make everything simple and easy, that's incredible!!!! congrats
@davidhitchen5369
@davidhitchen5369 2 жыл бұрын
Outstanding video. I knew you would assume that it wasn't perpendicular, but I couldn't come up with how to formulate the contradiction from that. My favorite historical proof by contradiction is Euclid's proof of the infinitude of primes.
@Sevinch_Jovliyeva
@Sevinch_Jovliyeva 2 жыл бұрын
you are the best teacher ever, thank you
@clintjoseph7126
@clintjoseph7126 6 жыл бұрын
This guy will get an OAM at some point in the future, I'm calling it now....
@ericBorja520
@ericBorja520 6 жыл бұрын
I gotta subscribe. I love your accent. You sound so fancy.
@razakhan6275
@razakhan6275 6 жыл бұрын
sir very nice iam big fan of you from india
@petebola2573
@petebola2573 4 жыл бұрын
Great video. Thought he was going to use Pythagoras Theorem after labelling points X and B. OA^2 = OX^2 + AX^2 OA^2 = (OB + BX)^2 + AX^2, but OB = OA therefore OA^2 = (OA + BX)^2 + AX^2 OA^2 = OA^2 + 2(OA)(OB) + OB^2 + AX^2 0= OA^2 + 2(OA)(OB) + OB^2 + AX^2 -2(OA)(OB) = OB^2 + AX^2 This is impossible because the sum of two squared number cannot be negative hence initial assumption must be wrong. Therefore tangent is always perpendicular to the radius of a circle it subtends. Question: If I started out by saying, "By Pythagoras Theorem, OA^2 is IDENTICAL to OX^2 + AX^2 (sorry, I can't use the three line equal sign) would line 3 be sufficient to support the contradiction ?
@malachitesardonyxvasquez
@malachitesardonyxvasquez 4 жыл бұрын
11:38 Can someone pls explain how he get Bx
@a_lampshade2278
@a_lampshade2278 4 жыл бұрын
Since the radius OB has an equal length to the radius OA, for OB + Bx to be less than OA, that would mean Bx would have to be less than zero.
@tldoesntlikebread
@tldoesntlikebread 4 жыл бұрын
I don't think it's O, I think he wrote Bx < 0
@cerenkoc.c
@cerenkoc.c 4 жыл бұрын
I don't know if you're Australian but your accent, sir, is impeccable. As a British I am intrigued :D Sounds much better than the American accent.
@sreeprakashneelakantan5051
@sreeprakashneelakantan5051 5 жыл бұрын
Second line you drew looked as if you used a ruler 👌
@siddharthvarshney5662
@siddharthvarshney5662 5 жыл бұрын
(I will use exactly the same figure as shown in the video) Instead of saying that Angle OAX < Angle OXA, we should make use of the properties of a Right Angled Triangle. We know that the side opposite to the right angle is the Hypotenuse (The Longest Side) So, OA > OB. The rest of the procedure is just the same. Is it correct to prove this theorem like this?
@meganlearns4395
@meganlearns4395 5 жыл бұрын
Hmm interesting, I may agree with you.
@MrGreenWhiteRedTulip
@MrGreenWhiteRedTulip 5 жыл бұрын
Siddharth Varshney yeah and honestly its better. You could also use the fact that sin,cos < 1
@theranpan_
@theranpan_ 5 жыл бұрын
Yes you could but the topic is of circles. So using of radii instead of sides of a triangles would be appropriate. Nevertheless your method is right too.
@advaitpetiwale9596
@advaitpetiwale9596 5 жыл бұрын
You can, Except there is one hole here. The triangle is OAX. Therefore, according to the hypotenuse is longest side property, it should be OA>OX*, so you basically still arrive at the same point he arrived. Only he gave them in even more detail cos it's easier for them to understand, and he's covering all questions that some curious minds among them will have 5 years later.
@ayli2843
@ayli2843 7 жыл бұрын
what math book do you suggest using that has challenging problems ?
@shreyasisrivastava9876
@shreyasisrivastava9876 6 жыл бұрын
Thanks ! This has helped me a lot.
@dannynamadula221
@dannynamadula221 4 жыл бұрын
This man is good. Life made easy
@rasheedoladimeji4092
@rasheedoladimeji4092 Жыл бұрын
You are wonderfully well, sir. However, I need a tutorial on domain and range of trigonometry functiond
@LEGONinjagoMB10
@LEGONinjagoMB10 Жыл бұрын
I continued with the supposition that BX can be negative which means that X is inside the circle but we know that X and A are in PQ and in the circle which means PQ isnt a tangent line because it intersects two points in a circle
@kevinbihari
@kevinbihari 3 жыл бұрын
Man, ox is one straight line
@tristanmoller9498
@tristanmoller9498 6 жыл бұрын
Amazing. Now what’s the square root of two?
@THEGLORYRISING
@THEGLORYRISING 5 жыл бұрын
Lets see if it can be written as a fraction, that is assume: sqrt(2)=p/q where p and q are whole numbers with no common factors Squaring both sides: 2=p^2/q^2 2q^2=p^2 So p^2 is divisible by 2. But any even square number must have a square root that is also even. This implies p is also divisible by 2 and can be written: p=2k where k is a whole number Now going back a few steps we can substitute p=2k for p: 2q^2=(2k)^2 2q^2=4k^2 q^2=2k^2 And we see q^2 is divisible by 2 implying q is divisible by 2 using the same reasoning as above. But then p and q each have a factor of 2 which is in contradiction with our assumption. So the square root of 2 can not be written as a fraction, so it is not rational. It must be irrational. QED
@shadow-cz3vr
@shadow-cz3vr 5 жыл бұрын
@@THEGLORYRISING Why can't p and q have common factors?
@THEGLORYRISING
@THEGLORYRISING 5 жыл бұрын
@@shadow-cz3vr We want p/q in its simplest form. If p/q has common factors it is not in its simplest form. In fact, we could then simplify p/q into a simpler fraction by dividing top and bottom by the greatest common factor (gcf) possible. If p and q have common factors then p=ba and q=ca (where a,b,c are natural numbers and a is the gcf) then ba/ca=b/c. We could then do the proof with b/c instead of p/q to achieve the same conclusion. If we don't state this, the proof doesn't work because the common factor we find may have been there to begin with. I should also add that the proof works by assuming something and then showing that the assumption creates an impossibility. We can assume whatever we want and in this case we are assuming p/q is in simplest form so that when we find that there's a common factor it is in contradiction with our assumption.
@chillmemer2525
@chillmemer2525 4 жыл бұрын
a surd
@winu1981
@winu1981 4 жыл бұрын
1.414 🤷🏻
@MohammadAli00
@MohammadAli00 2 жыл бұрын
Oh boy! I’m impressed of his straight line drawing ability
@carminefragione4710
@carminefragione4710 2 жыл бұрын
To explain these things you need a simple example. So , say you need to turn a ship around in the harbor, and for the distance you move forward into your turn, given the room available, how much of a change in direction or sharpness of the turn must you do to get the boat turned in the given space without crashing into the other side of the harbor. So for every yard you drive the boat, you need an angular turn that succeeds in getting the boat turned around, or 180 degrees, the direction is changed. So then that is why they used the sine and cosine, the perpendicular and the tangent, to navigate a ship in the water to turn about or bow , making it curve and turn, by having a way to calculate the steering , before you begin to turn the boat around.
@lifenote1943
@lifenote1943 3 жыл бұрын
what a BANGER video man
@RideR-SAM65
@RideR-SAM65 4 жыл бұрын
Sir can't we simply write if angle OxA = 90 the side opposite to this angle is hypotnuse that shoud be longer than other two sides... But it is clear from figure OA is less than Ox..it is only possible when x comes at A
@THETBONE
@THETBONE 5 жыл бұрын
thank you sir
@khanjarrrkhanjarrr4369
@khanjarrrkhanjarrr4369 4 жыл бұрын
1 question, shouldn't be tan 90 must be undetermined since you see it would be infinity/0 (Since perpendicular will be infinite and base would be 0 at tan 90 degree)
@simoncole4
@simoncole4 7 жыл бұрын
What grade are your students in?
@Carmine_Lupertazzi
@Carmine_Lupertazzi 5 жыл бұрын
D T mural
@MelonMediaMedia
@MelonMediaMedia 3 жыл бұрын
8:32 This itself is a proof by contradiction
@darshan5044
@darshan5044 2 жыл бұрын
Amazing , but I have an even easier explanation It takes into account the fact that the altitude is always the shortest path between a point and a given line on a plane. Here we take the tangent as the line, the radius as the point. As the tangent is just touching the circle, the shortest path is the radius (easy to see) hence it's perpendicular.
@satyapal8594
@satyapal8594 4 жыл бұрын
Excellent!
@aidangarner1181
@aidangarner1181 6 жыл бұрын
You can skip the bit about "smaller angles opposite smaller sides" just by saying that, since it's a right angle triangle, Pythagoras's theorem applies. Therefore c (the hypotenuse which is in this case OA) squared is equal to a squared plus b squared. Take a to be OX and b to be AX (although it doesn't matter which one is which) since you are adding b^2 to a^2 to make c^2. a (OX) must be less than c (OA). But this contradicts as OX = OB + BX and OB is equal to OA as they are both radii.
@davidadams5230
@davidadams5230 5 жыл бұрын
But you didn't explain how the tangent trig function relates to the tangent of a circle. Your didn't explain why the tangent function is called the tangent.
@rationalsceptic7634
@rationalsceptic7634 4 жыл бұрын
David Adams The Tangent cuts the Circle...I thought that was obvious?
@JohnSmith-rf1tx
@JohnSmith-rf1tx 4 жыл бұрын
That's the next video. He broke it up into two parts, and this was only part 1.
@jtm1283
@jtm1283 3 жыл бұрын
I think you omitted one fact that is needed for the formal proof: that a tangent to a circle is never inside the circle (at any other point). Without this X could be between B and O, instead of B being between O and X.
@shahk26
@shahk26 5 жыл бұрын
Can someone explain why sin Q (theta) = b/c and not b/a or c/a ? and same goes for cos (90-Q) /
@DaseAhrinia
@DaseAhrinia 5 жыл бұрын
Dont know if anyone replied to you about this... but how it goes is: by definition, Sine of an angle is equal to the ratio of "opposite side divided by the hypotenuse". Hence, sine Q would be b/c. by definition, Cosine of an angle is equal to the ratio of "adjacent side divided by the hypotenuse". Hence, cosine Q would be a/c. you can remember these ratios by using the mnemonic "SOHCAHTOA" which stands for "Sine = Opposite/Hypotenuse, Cosine = Adjacent/Hypotenuse, Tangent = Opposite/Adjacent"
@mrkitty777
@mrkitty777 4 жыл бұрын
Sos Cas Toa where s is hypothenusa, o is opposite side, a is near by side and S is Sinus, C is Cosinus, and T is Tangus soscastoa
@mrkitty777
@mrkitty777 4 жыл бұрын
Lol, previous commenter wrote it in English, didn't know it works in English too this little pony trick 😸😸
@AlexHeisEngholm
@AlexHeisEngholm 5 жыл бұрын
•˚• = therefor?
@doublex85
@doublex85 5 жыл бұрын
It's in Unicode. U+2234 ∴ Therefore. unicode-table.com/en/2234/
@blessos
@blessos 2 жыл бұрын
This is identical to another "proof" I just watched which is false/incomplete. Please identify the assumption which has been contradicted? It is not enough to draw a picture with the length Bx > 0, and then say because Bx < 0 we have a contradiction. What is the assumption which justifies the assertion that Bx > 0 as it appears in the picture?
@carlibouros
@carlibouros 2 жыл бұрын
Wow wow wow the bigger an angle, the bigger is his opposite side ? That is a very costly assumption, I'm gonna need more than "it is a triangles thing" ! By the way, a simplest demo is that the circle is symmetrical. So OAP = QAO, but OAP = 180° - QAO, so OAP = QAO = 90°
@kriskringleecommerce1367
@kriskringleecommerce1367 4 жыл бұрын
Very good!!!
@Celtics-x4w
@Celtics-x4w 6 жыл бұрын
So basically we could have just looked at line BX and automatically said that’s impossible
@aidangarner1181
@aidangarner1181 6 жыл бұрын
michael jordan nice proof
@colloidalsilver177
@colloidalsilver177 6 жыл бұрын
homie circles are pretty impressive
@yvesluyens5427
@yvesluyens5427 7 жыл бұрын
I am confused with the description of the angle: for me angle OAX is the right angle. I have always known that the letter in the middle is the angle or is there another convention? Thanks.
@Whizzer
@Whizzer 7 жыл бұрын
We say angle OXA is is 90 degrees, based on the assumption that OAX is smaller than 90 degrees. The drawing disagrees, so you should not look there, but instead follow the logic.
@yvesluyens9466
@yvesluyens9466 7 жыл бұрын
Whizzer191 Thank you. However it is obvious that it is a label mistake ; Mr Woo writes that OX is perpendicular to PQ so A and X should be reversed on the drawing. And writing angle MNP means angle N and nothing else. Best regards.
@mikejones8265
@mikejones8265 7 жыл бұрын
Yves, there are no mistakes in this video. Eddie is deliberately starting with a false situation in order to prove that it's false. We know that angle OAX is 90 degrees but only because we are told so. Eddie is proving that this is true by showing that it can't be any other size of angle. If you're still confused then keep watching the video until the penny drops. I'm a maths lecturer and this video is much better than other proofs I've seen on this particular topic.
@yvesluyens5427
@yvesluyens5427 7 жыл бұрын
Hi Mike, thank you for your answer. Wow, I made a real fool of myself here. Of course the video is correct. I listened to it the other day whilst doing something else and basically focused on the contradiction, ignoring the first six minutes. Eddie's videos are brilliant and inspiring. Cheers, Yves
5 жыл бұрын
Wait so why is tan called tan? 😭
@theranpan_
@theranpan_ 5 жыл бұрын
From tangent
@orlando5849
@orlando5849 4 жыл бұрын
tan is short for tangent. Tangent comes from Latin, tangere- to touch. A straight line can cut a circle at at two points of the circumference, In this case the line is called a secant ( from the Latin secare - to cut), or it can just touch at one point (tangent), or like the vast majority of lines miss altogether.
@jumpingjflash
@jumpingjflash 5 жыл бұрын
What class is this? If they've been learning maths for 12 years, what age are they?
@rationalsceptic7634
@rationalsceptic7634 4 жыл бұрын
jumpingjflash They are either 17 and Genii or he meant a smaller number!
@stevebennett7094
@stevebennett7094 Жыл бұрын
If you started watching this video at 6:50 you’d think this bloke was off his trolley 😅
@syatos
@syatos 5 жыл бұрын
You can also say that OA is in front of 90 deg so its the largest line in the riangle, but OX is outside the circle meanong its larger than the radius, and we have acontradiction again
@myday031
@myday031 4 жыл бұрын
Drawing straight line is just easy. They have mastered it like if you do it over and over again, you are also approaching to perfection but no one is perfect, just approaching.Assymptote? lol 😂😂 MATH IS ❤❤ Proving by Contradiction is kind'a hard. You always assume correct as incorrect. Therefore you imagine that incorrrect premise and interpret it in a logical way like its correct and still doesn't make sense at the end 😂😂.
@fishloh97
@fishloh97 5 жыл бұрын
Very detailed in explaining... But pace is way too slow for me. I m a quick and impatient learner ;>. Nonetheless you are a great teacher.
@JT-wk9sf
@JT-wk9sf 5 жыл бұрын
I don't understand why he says OAX is a cute angle. I actually find it rather unappealing...
@gyedublay
@gyedublay 4 жыл бұрын
It’s simply because you have a shallow understanding
@markwashington2412
@markwashington2412 4 жыл бұрын
wowwwwwww
@prabalsharma0111
@prabalsharma0111 4 жыл бұрын
7:11 done
@Martin-YouT
@Martin-YouT 2 жыл бұрын
QED.
@chupapimunyenyo1667
@chupapimunyenyo1667 2 жыл бұрын
FBI open the door!!!....Too smart
@harshjain196
@harshjain196 4 жыл бұрын
Will u teach me pleaseeeeeee
@TheRealBigBash
@TheRealBigBash 5 жыл бұрын
Skip the acute angle stuff and just go straight to the hypotenuse of a right angle triangle.
@sahelmohamed2650
@sahelmohamed2650 3 жыл бұрын
we must clone him....
@mrchin7562
@mrchin7562 4 жыл бұрын
I like
@JanBruunAndersen
@JanBruunAndersen 5 жыл бұрын
180 degrees in a triangle? Yeah, if it is a flat triangle. Try drawing that triangle on the surface of a ball and then measure the angles.
@andr101
@andr101 5 жыл бұрын
is he writing on a spherical desk?
@MC-mx1mt
@MC-mx1mt 5 жыл бұрын
lol of course we are assuming the triangle is drawn on flat euclidean geometry
@prakharpratapsingh5188
@prakharpratapsingh5188 5 жыл бұрын
Hey everyone, there is a basic mistake he did in his derivation, if anyone of you could find out it then i will get satisfied that I'm not only intelligent person in this comment section. Pls satisfy me.
@gyedublay
@gyedublay 4 жыл бұрын
Are you ok ?
@HypaNoob
@HypaNoob 3 жыл бұрын
Most annoying thing ive ever watched
Why are Tangent & Secant given their names?
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