😂😂👍Great fun!

@PreMath

2 ай бұрын

Excellent! Glad to hear that! Thanks ❤️

You have to state that angle ABC is a right angle otherwise triangle EBC and the constructed congruent triangle FBC do not give a straight line EBF which is necessary for you to apply angle bisector theorem on triangle CEF. Moreover, angle ABC being a right angle is necessary for use of tangent ratio in an alternative method for solution.

@waheisel

2 ай бұрын

Agree. I assumed that ABC was a right angle also. And I used the alternative solution you refer to; Used the angle bisector theorem to get length CB. Then calculated tanx and tan2x assuming B is a right angle. Then used tan sum formula to get AB.

Your solution implies that angle ABC is right angle which is not stated explicitly in the problem introduction

The tangent sum of angles formulas may be used to solve this problem. Let length BC = h. Then, tan(α) = 15/h, tan(2α) = 35/h and tan(3α) = (x + 35)/h. First, apply the tangent double angle formula tan(2α) = (2 tan(α))/(1 - tan²(α)): 35/h = (30/h)/(1 - (15/h)²). Solving for h, we get h² = 1575 and h = 15√7. Now the tangent sum of angles formula tan(α + ß) = (tan(α) + tan(ß))(1 - tan(α)tan(ß)) is used. Let ß = 2α, then tan(3α) = (15/h + 35/h))(1 - (15/h)(35/h)) = (50/h)/(1 - 525/h²). However, (x + 35)/h = tan(3α), so (x + 35)/h = (50/h)/(1 - 525h²) and x + 35 = 50/(1 - 525/h°) = 50/(1 - 525/1575) = 50/(1 - 1/3) = 50/(2/3) = 75. Since x + 35 = 75, x = 40 units, as PreMath also found.

@anthonycheng1765

2 ай бұрын

i use this approach.

@PreMath

2 ай бұрын

Thanks for sharing ❤️

@rabotaakk-nw9nm

2 ай бұрын

... h=15vʼ7; tanα=1/vʼ7 tan3α=(x+35)/(15vʼ7) tan3α=(3tanα-tan³α)/(1-3tan²α)= =(3/vʼ7-1/(7vʼ7))/(1-3/7)= =((21-1)/(7vʼ7))/(4/7)= =5/vʼ7=(x+35)/(15vʼ7) x+35=75; x=40...

@jimlocke9320

2 ай бұрын

Thanks for your interest in this method!

@ 3:17 the real dilemma for Schrodinger's Cat would have been thinking outside the box when confined in a triangle! 🤔🙂

@PreMath

2 ай бұрын

😀 Thanks ❤️

Thank you!

'Thinking outside the box' means 'Don' t be discouraged, there is an alternative way to solve the problem'. This is a great lesson for life. Great lesson, great prof. Thanks

Another super clear tutorial, thanks for sharing !

Let's do some trigonometrie: . .. ... .... ..... Since the triangles BCE, BCD and BCA are right triangles, we can conclude: tan(α) = BE/BC tan(2α) = BD/BC tan(3α) = AB/BC tan(2α) = 2tan(α)/[1 − tan²(α)] tan(2α)/tan(α) = 2/[1 − tan²(α)] (BD/BC)/(BE/BC) = 2/[1 − tan²(α)] BD/BE = 2/[1 − tan²(α)] (BE + DE)/BE = 2/[1 − tan²(α)] (15 + 20)/15 = 2/[1 − tan²(α)] 35/15 = 2/[1 − tan²(α)] 7/3 = 2/[1 − tan²(α)] 7/6 = 1/[1 − tan²(α)] 6/7 = 1 − tan²(α) 1/7 = tan²(α) ⇒ tan(α) = 1/√7 = √7/7 tan(2α) = 2tan(α)/[1 − tan²(α)] = 2*(√7/7)/(1 − 1/7) = (2√7/7)/(6/7) = (2√7/7)*(7/6) = √7/3 tan(3α) = [tan(α) + tan(2α)]/[1 − tan(α)tan(2α)] = (√7/7 + √7/3]/[1 − (√7/7)*(√7/3)] = (3√7/21 + 7√7/21)/(1 − 1/3) = (10√7/21)/(2/3) = (10√7/21)*(3/2) = (5√7/7) tan(3α)/tan(α) = (AB/BC)/(BE/BC) (5√7/7)/(√7/7) = AB/BE 5 = AB/BE ⇒ AB = 5*BE = 5*15 = 75 Finally we can calculate x: x = AD = AB − DE − BE = 75 − 20 − 15 = 40 Best regards from Germany

@unknownidentity2846

2 ай бұрын

Thinking outside the box can make things definitely easier. The method shown in the video is really nice.👍

@PreMath

2 ай бұрын

Excellent! Thanks for sharing ❤️

@User-jr7vf

2 ай бұрын

No, you shouldn't assume those are right triangles. PreMath didn't state that in the introduction, and he gets the right answer without making that assumption.

@Birol731

2 ай бұрын

Mein Lösungsvorschlag, fand ich auch am leichtesten.....🙏👍

@VictorLonmo

2 ай бұрын

This is how I did it too.

In summary, using trigonometry: Let BC = y. Then tan⁻¹(35/y) = 2 tan⁻¹(15/y) ⇒ y = 15√7 ⇒ tan(α) = √7/7 ⇒ tan(2α) = √7/3 ⇒ tan(3α) = 5√7/7 so AB = 75 ⇒ AD = x = 40. PreMath's solution is really elegant! Thank you very much! 🙏

@PreMath

2 ай бұрын

Excellent! You are very welcome! Thanks for sharing ❤️

Great exercise

"Think outside the box" States that 'we have to use 200% of our brain'

@PreMath

2 ай бұрын

😀 Thanks ❤️

My way of solution is ▶ First check tan(α) tan(α)= EB/CB CB= y EB= 15 ⇒ tan(α)= 15/y tan(2α)= DB/CB DB= 20+15 DB= 35 ⇒ tan(2α)= 35/y Let's write tan(2α) in form of tan(α): tan(2α)= 2*tan(α)/[1-tan²(α)] ⇒ 35/y= 2*(15/y)/[1-(15/y)²] 35/y= (30/y)/[1- 225/y²] 35/y= (30/y)/[(y²-225)/y²] 35*(y²-225)= 30y 35y²-7875 = 30y 5y²= 7875 y²= 1575 1575= 25*63= 25*7*9 ⇒ y²= 5²*3²*7 y= 15√7 tan(3α)= AB/CB AB= 35+x CB= y CB= 15√7 tan(3α)= (35+x)/15√7 Let's write tan(3α) in form of tan(α) and tan(2α): tan(3α)= [tan(α)+tan(2α)]/[1-tan(α)*tan(2α)] tan(α)= EB/CB ⬆ tan(α)= 15/15√7 tan(α)= 1/√7 tan(α)= √7/7 tan(2α)= DB/CB ⬆ tan(2α)=35/15√7 tan(2α)=7/3√7 tan(2α)= √7/3 ⇒ (35+x)/15√7 = [√7/7 + √7/3]/[1- (√7/7)*(√7/3)] (35+x)/15√7 = [(3√7+7√7)/21]/[(21-7)/21] (35+x)*(14/21)= 15*√7*10√7/21 (35+x)*14= 150*7 (35+x)*2= 150 35+x= 75 x= 75-35 x= 40 units lengths ✅

@PreMath

2 ай бұрын

Thanks for sharing ❤️ .

@robertlynch7520

2 ай бұрын

Which (while I didn't get involved with all the √(7) parts) was exactly the same technique I worked through … in decimal. Algebraically. [1.1]  𝒕 = tan(θ) [1.2]  𝒕 = √(1 ÷ 7) this was a longer algebraic step than shown [1.3]  𝒕 = 0.377964 [2.1]  𝒈 = tan(2θ) [2.2]  𝒈 = 2𝒕/(1 - 𝒕²) substitute in [1.3] and solve [2.3]  𝒈 = 0.881916 [3.1]  𝒌 = tan(θ ⊕ 2θ) [3.2]  𝒌 = (𝒕 + 𝒈) / (1 - 𝒕𝒈) same ... [1.3] and [2.3] now [3.3]  𝒌 = 1.889818 Then, armed with those tangents [4.1]  H = height of BC [4.2]  H = 15 / 𝒕 [4.3]  H = 15 ÷ 0.377964 good ol' [1.3] substitution [4.4]  H = 39.68634 And extension (where [L₁ = 15], [L₂ = (15 + 20)] and [L₃ = (15 + 20 + 𝒙)]: [5.1]  L₃ = H⋅𝒌 … substitute [4.4] and [3.3] [5.2]  L₃ = 39.68634 × 1.889818 [5.3]  L₃ = 75 [5.4]  (15 + 20 + 𝒙) = 75 … incorporate concept [5.5]  (35 + 𝒙) = 75 … rearrange [5.6]  𝒙 = 75 - 35 [5.7]  𝒙 = 40 Ta Da. ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅

At 0:57, Use Angle-Bisector Theorem on triangle CDB, Let CB=h and CE is the angle bisector. CD/CB=DE/EB CD/h= 20/15=4/3 CD = 4h/3. CD^2=DB^2+CB^2 Pythagoras. (4h/3)^2=(20+15)^2+h^2 (16h^2)/9=35^2+h^2 16*h^2 =9*(35)^2+9*h^2)+ 7*h^2 = 9*(35^2) h^2=9*(35^2)/7 h = (3*5*7)/sqrt7 =15*sqrt7 units. tan(α)= 15/h = 15/(15*sqrt7) =1/sqrt7..................................(1) tan(2α)= DB/CB =35/h = 35/(15*sqrt7)=7/(3*sqrt7)............(2) Using the relationship tan(A+B)= (tanA+tanB)/(1-tanA*tanB) Let A=(2α) and B=(α), tan(2α+α)=(tan(2α)+tan(α))/(1-tan(2α)*tan(α)) tan(3α) = (7/(3*sqrt7)+1/sqrt7)/(1-7/(3*sqt7)*1/(sqrt7)= (10*sqrt7)/21)/(2/3)=5/sqrt7 tan(3α) = 5/sqrt7......................................................................(3) Consider triangle ABC, tan(3α)= 5/sqrt7=AB/CB=AB/h = (x+20+15)/(15*sqrt7) (x+35)=(5/sqrt7)*(15*sqrt7)=75 x= 75-35 =40 units. Thanks for the puzzle professor.

sistema usando pitágoras e teorema da bissetriz interna. parabéns, obrigado! seu jeito foi mais elegante.

What do we have here? 1) BC = K (constant) 2) tan(1*alpha) = 15/K ; K = 15/tan(alpha) 3) tan(2*alpha) = 15/K ; K = 35/tan(2*alpha) 4) tan(3*alpha) = (35 + X)/K ; K = (35 + X)/tan(3*alpha) So: 5) 15/tan(alpha) = 35/tan(2*alpha) 6) 15/tan(alpha) = (35 + X)/tan(3*alpha) 7) 35/tan(2*alpha) = (35 + X)/tan(3*alpha) 8) tan(2*alpha) = [2*tan(alpha)] / [(1 − tan^2(2*alpha)] 9) These Equality, I think (therefore I am), are the Mathematical Tools I need to solve this Problem. I will return soon. Now I must go out.

@PreMath

2 ай бұрын

Thanks for sharing ❤️

1) 15/tan(a) = 35/tan(2a) ; tan(2a) = 35tan(a)/15 ; tan(2a) = 7tan(a)/3 2) tan(2a) = 7tan(a)/3 3) tan(2a) = 2tan(a) / (1 − tan^2(a)) 4) 7tan(a)/3 = 2tan(a) / (1 − tan^2(a)) 5) 7tan(a) * (1 - tan^2(a)) = 6tan(a) 6) 7tan(a) - 7tan^3(a) = 6tan(a) 7) 7tan(a) - 6tan(a) - 7tan^3(a) = 0 8) tan(a) - 7tan^3(a) = 0 9) tan(a) = Y 10) Y - 7Y^3 = 0 11) Y = -sqrt(7)/7 ; Y = 0 ; Y = sqrt(7)/7 12) tan(a) = sqrt(7)/7 ~ 0,37796 13) arctan(sqrt(7)/7) ~ 20,7º 14) Angle (a) = 20,7º 14) BC ~ 39,6863 lin un 15) Angle (3a) = 62,1144º 16) tan(3a) = AB / BC ; 1,8898 = AB / 39,6863 ; AB = 1,8898 * 39,6863 ; AB ~ 74,99995 lin un 17) AD = AB - 35 ; AD = 74,99995 - 35 = 39,99995 lin un 18) My answer is that AD = X = 40 Linear Units.

شكرا لكم على المجهودات البداية ABC قائم الزاوية فيB يمكن استعمال tan3a, tan2a,tana .... a=ECB BC=d tana=15/d tan2a=35/d d=15(جذر5) tana=(جذر7)/7 tan2a=(جذر7)/3 tan3a=5(جذر7)/7 35+x/15(جذر7) = tan3a x=40

Is there any short book or short paper listed on the internet which summarises all the theorems you demonstrate in your examples?

Io ho utilizzato sempre il teorema dei seni,da cui risultano (cosα)^2=7/8,x=40

@PreMath

2 ай бұрын

Excellent! Thanks for sharing ❤️

(20+x)/(15×2)=x/20, 30x=20(20+x)=400+20x, 10x=400, x=40.😊

@PreMath

2 ай бұрын

Excellent! Thanks for sharing ❤️

I propose a trigonometric solution (it is less simple than yours, so less good). I note t in place of alpha We have tan(t) = 15/BC, tan(2.t) = 35/BC , tan(3.t) = (35 +x)/BC Tet's calculate BC first. We know thatv tan(2.t) = (2.tan(t))/ (1 - tan(t)^2), so tan(2.t) = (30/BC)/ (1 - (225/BC^2)) = (30.BC)/ (BC^2 - 225) So we have: 35/BC = (30.BC)/ (BC^2 -225) and then 30.BC^2 = 35.BC^2 - 225.35. That gives BC^2 = 1575 and BC = 15.sqrt(7) Now we know that tan(t) = 15/(15.sqrt(7)) = sqrt(7)/7 and tan(2.t) = 35/(15.sqrt(7) = sqrt(7)/3 We know that tan(3.t) = (tan(t) + tan(2.t)) / (1 - tan(t).tan(2.t)), so tan(3.t) = (sqrt(7)/7) + sqrt(7)/3) / (1 - (sqrt(7)/7).(sqrt(7)/3)) tan(3.t)= ((10.sqrt(7))/21) / (2/3) = (5.sqrt(7))/21. And so, finally (5.sqrt(7))/ 21 = (35 +x)/ BC = (35 +x)/ (15sqrt(7)) 5.sqrt(7).15.sqrt(7) = 7. (35 +x) and 75 = 35 +x, giving x = 40 at the end.

@PreMath

2 ай бұрын

Thanks for sharing ❤️

❤❤❤ thanks dear teacher may God bless you 💯💯💯

@PreMath

2 ай бұрын

You are very welcome! Thanks dear ❤️ You are so generous🌹

I got tan alpha and tan 2 alpha and let tan alpha= t. I then got (35-35t sq)/2t =15/t That gave me t=1/ sqroot7. Then I expanded tan 3 alpha and substituted for t and put result = x+35/15(root7) getting x=40.

Is this a right-angled triangle? There is no mention on this.

@User-jr7vf

2 ай бұрын

It is, but you should not assume that a priori. As he shows, there is a way to get the right answer without making that assumption.

@hongningsuen1348

2 ай бұрын

You are right that angle ABC has to be a right-angle. If it is not a right angle, the construction of congruent triangles EBC and FBC will lead to line EBF not being a straight line and subsequently ECFB is a quadrilateral, not a triangle that is used by Premath for application of angle bisector theorem.

I did it with tangents. t = tan(alpha) = tan(a) where a = alpha t = 15/h, ht = 15 tan(2a) = 35/h = 2t/(1 - t^2), 2ht = 35(1 - t^2) = 30, 1 - t^2 = 30/35, t^2 = 5/35 = 1/7, t = 1/sqrt(7) tan(3a) = (35 + x)/h = (tan(a) + tan(2a))/(1 - tan(a)tan(2a)) = (t + 2t/(1 - t^2))/(1 - t.2t/(1 - t^2)) = (3t - t^3)/(1 - 3t^2) = (1/sqrt(7))*(3 - 1/7)/(1 - 3/7) = (1/sqrt(7))*20/4 = 5/sqrt(7) = (35 + x)/h Now h = 15/t = 15sqrt(7), (35+x)/h = (35+x)t/15 = (35+x)/(15sqrt(7)) = 5/sqrt(7) therefore 35 + x = 5*15 = 75, x = 40 And after writing this, I see that many others used the same method.

Thanks for the video but for me it makes things a little easy to understand/imprint when the example picture is close to the dimensions and angles of the calculations. The X in the pic isn't close to 40 in length unless it's an optical illusion. Just checked and thankfully for me my vision isn't screwed, the X along the brown triangle isn't 40 units in the pic.

Hello! I used tan(2x)=(2×tan(x))/(1-tan²(x)) But I think, in both cases we need angle ABC to be a right angle, don't we? Greetings!

X=40, may be Angle ACE, CD is the bisector, So AC/X=CE /20 20AC=X.CE Let AC=h CD=m CE =n CB =p 20h=xn......... E1 Angle CDB, CE is the bisector, So CD/DE=CB/EB m/20=p/15 15m=20p P=3m/4.......... E2 In triangle CEB n^2=p^2+225........ E3 In triangle CDB m^2=p^2+(35)^2 =p^2+1225 =(3m/4)^2+1225 ={(9m^2)/16}+1225 =(9m^2+19600)/16 16m^2=9m^2+19600 7m^2=19600 m^2=19600/7=2800 m=20.*7(*=read as root ) In E2, P=3m/4 P=(3×20.*7)/4=15.*7 In triangle BEC, n^2=p^2+(15)^2 =(15.*7)^2+225 =1575+225=1800 n=30.*2 In triangle ABC, h^2=p^2+(35+X)^2 =1225+x^2+70x+1575(p^2=1575) =2800+x^2+70x As per E1, 20h=x. n, so h=x. n/20 =(x.30.*2)/20 h^2=(x^2.1800)/400 =18x^2/4 (18x^2)/4=2800+x^2+70x 18x^2=11200+4x^2+280x 14x^2-280x-11200=0 14(x^2-20x-800)=0 X^2-20x-800=0 X^2-40x+20x-800=0 X(x-40)+20(x-40)=0 (X-40)(x+20)=0 X=40 or x= (-20) X=(-20)is not accepted (length is not negetive ) So, x=40

If you can find a simple value for BC then you have found a solution to an age old geometric problem that has evaded a solution - how to trisect an angle. This problem only works for one value of BC - what is it?

tan(α) = 15/h tan(2α) = 35/h tan(3α) = (x+35)/h tan(2x) = 2tan(x)/(1-tan²(x)) tan(2α) = 2tan(α)/(1-tan²(α)) 35/h = 2(15/h)/(1-(15/h)²) 35/h = (30/h)/(1-(225/h²)) 35(1-(225/h²)) = h(30/h) 35 - 225(35)/h² = 30 (35h²-7875)/h² = 30 35h² - 7875 = 30h² 5h² = 7875 h² = 1575 h = √1575 = 15√7 tan(x+y) = (tan(x)+tan(y))/(1-tan(x)tan(y)) tan(2α+α) = (tan(2α)+tan(α))/(1-tan(2α)tan(α)) tan(3α) = ((35/15√7)+(15/15√7))/(1-(35/15√7)(15/15√7)) (x+35)/15√7 = (50/15√7)/(1-(525/1575)) x + 35 = 50/(1-1/3) x + 35 = 50(3/2) x + 35 = 75 x = 75 - 35 = 40

Sometimes you need to think instead of applying pattern actions. Cross multiplying equal fractions is not always the best way. X/2=(X+20)/3 easy to see that is it X/6=20/3 so X=40 ended up not cross multiplying at all. (If not so easy to see subtract X/3 from both sides). Although extra construction outside of big triangle is clever yea. Other trig methods in comments are not optimal, easier for machines but for humans author’s solution is better I think.

Shouldn’t we have x/20=20/15? Maybe not.

А дальше? 15;15;20;40;?? Дальше 240😅

(20)^2=400 (15)^2=225 {400+225}=625 3(15°)=45° 3(15°)=45° {45°+45°+90°)=180° {625\180°=3.√85 3.5^√17 3.5^√17^1 3.5√1^√1 3 5 (ABCDF+3ABCDF-5)