You are the best, i told you already. I am italian, but i prefer to follow you that explain in english instead to follow italian youtubers. Are you a math professor? Anyway gratz and thank you for your lessons

@PreMath

11 ай бұрын

Wow, thank you! You are awesome, Andrea. Keep smiling👍 Love and prayers from the USA! 😀 Yes, I'm an educator!

I used a slightly different approach. In △ACD, we can use Pythagorean theorem to find CD = √(x² − 1) In △BCD, we use Law of Sines to get: BD/sin(∠BCD) = CD/sin(∠B) 1/sin(30°) = √(x² − 1)/sin(∠B) sin(∠B) = sin(30°) √(x² − 1) sin(∠B) = (1/2) √(x² − 1) In △ABC, we use Law of Sines to get: AB/sin(∠ACB) = AC/sin(∠B) (x+1)/sin(120°) = 1/sin(∠B) sin(∠B) = sin(120°)/(x+1) sin(∠B) = (√3/2) / (x+1) Now we set both values of sin(∠B) equal to each other (1/2) √(x² − 1) = (√3/2) / (x+1) √(x² − 1) = √3 / (x+1) Square both sides: (x² − 1) = 3 / (x² + 2x + 1) (x² − 1) (x² + 2x + 1) = 3 x⁴ + 2x³ + x² − x² − 2x − 1 = 3 x⁴ + 2x³ − 2x − 4 = 0 x³ (x + 2) − 2 (x + 2) = 0 (x + 2) (x³ − 2) = 0 Possible solutions: x = −2 → not possible, since x is a side length x³ = 2 *x = ∛2*

@PreMath

11 ай бұрын

Thanks for watching!

@timeonly1401

11 ай бұрын

My solution as well. (I wasn't as direct as yours.. Did some extra work that wasn't needed) got the same quartic polynomial, and was pleasantly surprised that it easily factored by grouping. Are you as surprised as I am that out of a simple geometric situation sprang a cube root?! 😅

تمرين جيد جميل. رسم واضح مرتب. شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويحميكم ويرعاكم جميعا . تحياتنا لكم من غزة فلسطين

@PreMath

10 ай бұрын

Thanks dear🙏

@ 9:22 we all start tweakin! 🙂

@PreMath

11 ай бұрын

Thanks for watching!

That's just freaked me out. I'm sure this is magic 🤓👍🏻

@PreMath

11 ай бұрын

Thanks for watching!

This is the real stuff, many thanks, Sir! You made it!

@PreMath

11 ай бұрын

Glad you liked it! Thanks for watching!

It's absolutely stunning to me that such a "simple" geometrical situation can result in a CUBE ROOT!! Amazing! Thx for this. 😮 🎉

A fascinating and unexpected result. Because 2^(1/3) is not constructible with compass and straight edge, the entire figure as shown is not constructible.

@murdock5537

11 ай бұрын

CAB ≈ 37,47° → ABC ≈ 22,53° - let's try...

@PreMath

11 ай бұрын

Thanks for watching!

@valentinaivanova4806

11 ай бұрын

You can build a drawing that reflects real proportions: 1. Build a DCB angle with infinite beams 2. Lay an AC beam 3. Apply the ruler, constantly monitoring the lengths of AC and DB segments until they are approximately equal.

My solution was all based on geometry: First idea how to start solving this was looking at the angle 30 degrees, should I construct some angle of 90 or 60 degrees out of that, which I also did: First I made a rotated copy of triangle CBD, so that the new triangle BEC forms a trapezoid with CBD, and the edge BE is parallel to DC, and CE parallel to AB. Next, looking at angles CAD = theta, and ABE = 90-theta, so that if I extend the lines AC and BE into a new point F, the angle AFB is a right angle. Therefore, the large triangle ABF and the smaller one CEF are uniform having the same angles, and proportional lengths of edges. Next, the triangles BCF is a right angle triangle having also a 60 degrees angle at point C, a triangle having lengths with proportions (1, sqrt(3), 2). The lengths of the large right triangle were AF = 1+1/x, BF = sqrt(3)/x, and the hypothenyse AB = x + 1, so applying the Pythogorean gave (1+1/x)^2 + (sqrt(3)/x)^2 = (x+1)^2 and solving that resulted with x = cuberoot(2). (I probably should have made a video out of this solution, instead of writing this as a text that might be really hard for anyone to get the idea)

@PreMath

11 ай бұрын

Super! Thanks for sharing👍

Absolutely beautiful! 🤩

@PreMath

11 ай бұрын

Thank you! Cheers!

Amazing video👍 Thank you so much for your hard work 😊

@PreMath

11 ай бұрын

Thanks for watching!

Convoluted but very worthwhile!😀❤🥂

@PreMath

11 ай бұрын

Thanks for watching!

Well done.

@PreMath

11 ай бұрын

Thanks for watching!

Thanku brother

@PreMath

11 ай бұрын

Thanks for watching!

tq dear professor ❤

@PreMath

11 ай бұрын

You are welcome, Rishu You are awesome. Keep smiling👍 Love and prayers from the USA! 😀

Construct a line through point B parallel to CD and extend AC to meet that line. Call the intersection point F. Note that

@PreMath

11 ай бұрын

Thanks for watching!

@jimlocke9320

11 ай бұрын

@@PreMath You're most welcome and thank you for producing the video! Here's another way to the solution after we have found BF = √3/x. From the Pythagorean theorem, CD = √(x²-1), thus CD/BF = (√(x²-1))/(√3/x). From similar triangles, CD/BF = AD/AB and we have AD/AB = x/(x+1), so (√(x²-1))/(√3/x) = x/(x+1). Multiplying both sides by (√3/x), we get √(x²-1) = (√3/x)( x/(x+1)), which simplifies to √(x²-1) = √3/(x+1). Square both sides and we get x²-1 = 3/(x+1)² or x²-1 = 3/(x²+2x+1). Multiply both sides by (x²+2x+1) and simplify: x⁴ +2x³-2x-4= 0. Factor out (x+2) to get (x+2)(x³)-(x+2)(2)= 0 or (x+2)(x³-2)= 0. The 2 solutions are x = -2 and x³ = 2. The value of x must be positive, so x = -2 is invalid and x = ³√2, as PreMath found.

Feels very difficult. 😗First cos A is 1/x, then BC^2=1+(1+x)^2-2(1+x)/x, DE=sqrt(x^2-1)/2, CE=sqrt(x^2-1)sqrt(3)/2, EB=sqrt(1-(x^2-1)/4)=sqrt((5-x^2)/4), so BC=CE+EB=a clumsy expression of x, so I cannot continue my work.😢

@PreMath

11 ай бұрын

Thanks for watching!

You could avoid fractional denominators by using sec (Hypotenuse ÷ Adjacent) and csc (Hypotenuse ÷ Opposite). The Law of Sines can ironically be expressed as a csc(α) = b csc(β).

@PreMath

11 ай бұрын

Thanks for watching!

cut out ACD and rotate anticlockrise until AC is the bace line and put it on BDC, notice they have two same side, rotate the line DC from ACD anticlockrise can make the triangle BDC, therefore it should be a formula to get the information of x.

@PreMath

11 ай бұрын

Thanks for watching!

You are the best in KZread but x=2 is one of the solutions

@timeonly1401

11 ай бұрын

Actually, the other "solution" is x = -2 , which is not allowed because x is a length of a line segment, so it must be true that x > 0 (which it ain't, if x = - 2 ! 😊 ) .

Line A B is 28m.m.form point B is tilted at C, angel ABC is 150° distance from AC is 34m.m what's the length of BC? Can you solve it please

X = ad+ db = x os 2

I thought I had solved this, but my reasoning was wrong. I'm not even going to watch the video to find out how I should have done it, though, because you have deliberately tried to fool us with the diagram. If a triangle is not isosceles, it should not be drawn as if it might be. I don't like that sort of trickery. Similarly, if an angle is not a right angle, it should not be drawn in such a way that it is near-indistinguishable from one. The fact is that triangle CDB _looks_ isosceles, yet it can't be, because triangle ACD would then have to be an isosceles right triangle, which, in turn, would be incompatible with triangle CDB if it were isosceles. I know you always say that the diagram may not be 100% to scale, but here have tried to run us off the rails from the start. I hope this is not the beginning of a new trend.

@PreMath

11 ай бұрын

Thanks for watching!

@WernHerr

11 ай бұрын

I do not think so. Sides of the same length are clearly marked as such, otherwise I assume that they are not of the same length.

@AnonimityAssured

11 ай бұрын

@@WernHerr Yes, I was just angry with myself, really, for not being able to do it. I still can't do, but I can't face watching the video to find out what I'm missing. It will just have to be a question that I _can_ afford to ignore. Be aware, though, that congruent or similar features are _not_ always indicated in these problems. Sometimes, the answer depends on determining that two given lengths are equal, or that two triangles are similar.

@valentinaivanova4806

11 ай бұрын

​​​@@AnonimityAssuredYou can build a drawing that reflects real proportions: 1. Build a DCB angle 30 grad with infinite beams 2. Lay an AC beam 3. Apply the ruler, constantly monitoring the lengths of AC and DB segments until they are approximately equal.

@AnonimityAssured

11 ай бұрын

@@valentinaivanova4806 Thanks. I found a way of constructing a reasonably accurate diagram within the limits of MS Word. I was right that triangle CDB should be significantly more asymmetrical, with CD a bit less than 4/5 of DB. Now, having finally watched the video, I see that I would never have solved the problem. I'll just have to archive it under the label 'Techniques I should know, but never will.'

A me risulta x=rad((1+rad13)/2)...dal teorema dei seni...x+1:sin120=1:sin(120+arccos1/x)...no!ho sbagliato...risulta ,dal teorema dei seni..x^4+2x^3-2x-4=0...che da soluzione positiva x=rad3(2)...spero sia giista

@PreMath

11 ай бұрын

Thanks for watching!

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