Very good solution!! We have to think outside the box.

@PreMath

Ай бұрын

Yes, exactly Thanks for the feedback ❤️

Thank you! Your method is a lot better than mine: AE= 40- 20sqrt 3= 20(2-sqrt3)--> tan (x) = 20/ 20(2-sqrt3)= 1/(2-sqrt3) --> x= 75 degrees😅

@marcelowanderleycorreia8876

Ай бұрын

I did the very same way you posted... It´s right, but the way that the teacher solved, is something.... We have to think outside the box...

@PreMath

Ай бұрын

Excellent! Thanks for the feedback ❤️

@PreMath

Ай бұрын

You are very welcome! Thanks for sharing ❤️

Thank you for the proof

@PreMath

Ай бұрын

You are very welcome! Thanks ❤️

You made it so easy again. Great idea!

In ∆ BCE Tan(60)=BE/BC BE=BCtan(60°)=20√3 AE=40-20√3 In ∆ ABF Tan(x)=AD/DF=20/(40-20√3) So x=75°.❤❤❤ Thanks sir.

@PreMath

Ай бұрын

Excellent! You are very welcome! Thanks for sharing ❤️

Thank you!

@PreMath

Ай бұрын

You're welcome!❤️

I did not catch both legs being 40. I should have known you would have a sneaky way of doing it. 👍

@PreMath

Ай бұрын

Thanks for the feedback ❤️

Thanks Sir That’s nice With glades ❤❤❤❤

ΔBCEE is a special 30°-60°-90° right triangle, so its long side, BE, is √3 times as long as its short side, or 20√3. AE = AE - BE = 40 - 20√3 = 20(2 - √3). For ΔAEF, side EF = 20 and AE/EF = (20(2 - √3))/20 = 2 - √3. We can multiply by (2 + √3)/(2 + √3) and simplify, to find that the long side is (2 + √3) times as long as the short side, which is one way of expressing the ratio of sides for a 15°-75°-90° right triangle. So, angle

@PreMath

Ай бұрын

Excellent! Thanks for sharing ❤️

I made the solution too complex again. I expected it to be 15 degrees less than 90, so I bisected angle BEC forming two 15 degree angles then using the half angle theorem got the sine of the 15 degree angle to be 2-sqrt3. One can easily show that AEF also has sine 2-sqrt3 so x is 90-15=75. Thanks for the exciting daily puzzle!

@PreMath

Ай бұрын

You are very welcome! Thanks for sharing ❤️

Yes, got it

@PreMath

Ай бұрын

Excellent! Thanks for the feedback ❤️

It is very easy. In triangle CBE: EB/BC = tan(60°) = sqrt(3), so EB = BC.sqrt(3) = 20.sqrt(3). Then AE = AB - EB = 40 -20.sqrt(3), so DF = AE = 40 - 20.sqrt(3) In triangle ADF: tan(x) = AD/DF = 20/(40 -20.sqrt(3)) = 1/(2 -sqrt(3)) = 2 + sqrt(3) = cotan(15°) = tan(90° -15°) = tan(75°). So x = 75° and that's all.

@PreMath

Ай бұрын

Thanks for sharing ❤️

Before viewing the video: I might call EB 20*sqrt(3) due to the properties of a 30,60,90 triangle. Therefore, AE is 40 - 20*sqrt(3). DF is that length too. AD = 20 (given) and DF is 40-(20*sqrt(3)) Angle x is tan(-1)(20/(40-20*sqrt(3))) This can be reduced to tan(-1)(1/(2-sqrt(3)) tan(-1)(1/(2-sqrt(3))) is 75 degrees according to my calculator, but I imagine the video will show a better way.

@PreMath

Ай бұрын

Thanks for sharing ❤️

Why we use alpha in this geometry?? Plz answer😥😥

As EFCB is a rectangle, all internal angles are 90° and opposite sides are parallel, so BE = FC and CB = EF. Let ∠BEC = α. As ∠ECB = 60° and ∆CBE is a right triangle, α = 30°. tan(α) = CB/BE 1/√3 = 20/BE BE = 20√3 EA = 40 - 20√3 = 20(2-√3) As ADFE is a rectangle, all internal angles are 90° and opposite sides are parallel, so EA = DF and FE = AD. tan(x) = AD/DF tan(x) = 20/20(2-√3) tan(x) = (2+√3)/(2-√3)(2+√3) tan(x) = (2+√3)/4-3 tan(x) = 2 + √3 x = tan⁻¹(2+√3) = 75°

@PreMath

Ай бұрын

Excellent! Thanks for sharing ❤️

FB=40, angle ABF= 30, angle FAB=angle X= (180-30)/2=75!

@PreMath

Ай бұрын

Excellent! Thanks for sharing ❤️

Tg (X)=20/(40-20√3)=2+√3 ; X=75° Gracias y saludos.

@PreMath

Ай бұрын

Excellent! Thanks for sharing ❤️

1) Triangle [EBC] = Triangle [EFC] ; they are both (30º ; 60º ; 90º) 2) tan(60º) = EB / 20 ; EB = 20 * tan(60º) ~ 34,64 lin un 3) AE = DF = 40 - (20 * tan(60º)) 4) tan(X) = AD / DF 5) tan(X) = 20 / [40 - 20 * tan(60º)] 6) tan(X) = 20 / [20 * (2 - tan(60º)] 7) tan(X) = 1 / (2 - tan(60º)) 8) X = arctan(1 / (2 - tan(60º)) 9) X = 75º 10) Answer : The Angle X is equal to 75º.

@PreMath

Ай бұрын

Excellent! Thanks for sharing ❤️

Wow so much easier than invtan (2+√3)

@PreMath

Ай бұрын

Glad to hear that! Thanks for the feedback ❤️

tgx=20/(40-20tg60)=1/(2-√3)=2+√3

@PreMath

Ай бұрын

Thanks for sharing ❤️

Let's find x: . .. ... .... ..... The triangle BCE is a right triangle, so we can conclude: tan(∠BCE) = BE/BC ⇒ BE = BC*tan(∠BCE) = 20*tan(60°) = 20*√3 AE = AB − BE = 40 − 20*√3 = 20*(2 − √3) tan(∠AFD) = AD/DF = BC/AE tan(x) = 20/[20*(2 − √3)] = 1/(2 − √3) = (2 + √3)/[(2 − √3)(2 + √3)] = (2 + √3)/(4 − 3) = (2 + √3) ⇒ x = 75° Here is the proof: tan(75°) = tan(45° + 30°) = [tan(45°) + tan(30°)]/[1 − tan(45°)tan(30°)] = (1 + 1/√3]/(1 − 1*1/√3) = (√3 + 1)/(√3 − 1) = (√3 + 1)²/[(√3 − 1)(√3 + 1)] = (3 + 2√3 + 1)/(3 − 1) = (4 + 2√3)/2 = 2 + √3

@robertlynch7520

Ай бұрын

I like it. However, I ended up solving this by using the identity [1.1]  tan 2θ = (2 tan θ) / (1 - tan² θ) … and renaming variables as [1.2]  D = 2H / (1 - H²) … D is "double" and T is "tangent" By itself, this doesn't solve the problem. But I got to wondering, "what is the tangent of 15°? " (guessing that 15° would figure into the solution). Well, substituting [θ = 15°] for [1.1] [2.1]  tan 2•15° = 2 tan 15° / (1 - tan² 15°) [2.2]  tan 30° = 2H / (1 - H²) … and solving for H [2.3]  1 / √3 = 2H / (1 - H²) … rearranging [2.4]  1 - H² = 2√3 H … and moving stuff into quadratic form [2.5]  0 = 2√3 H + H² - 1 … then solving for H (work…) [2.6]  H = (-1 ±√(1 + 1/√3²)) / (1/√3) [2.7]  H = 2 - √3 [2.8]  tan 15° = 2 - √3 Well, how about that. Looking back at the original diagram, what do we have? [3.1]  BC = 20 × 1 [3.2]  BE = 20 × √(3) [3.3]  BA = 20 × 2 … and thus [3.4]  EA = 20 × (2 - √(3)) So the ∠AFE must have tangent of 20 × (2 - √(3)) ÷ 20 × 1 → (2 - √(3)), which is 15° Cool. 𝒙 is the complementary angle, 90° - 15° = 75° And done. Yay. ⋅-=≡ GoatGuy ✓ ≡=-⋅

@PreMath

Ай бұрын

Awesome! Very much appreciated! Thanks for sharing ❤️

@unknownidentity2846

20 күн бұрын

@@robertlynch7520 Your approach is also very nice and I used it already in the past. However, in this case the following method is easier: tan(15°) = tan(45° − 30°) = [tan(45°) − tan(30°)]/[1 + tan(45°)tan(30°)] = (1 − 1/√3]/(1 + 1*1/√3) = (√3 − 1)/(√3 + 1) = (√3 − 1)²/[(√3 + 1)(√3 − 1)] = (3 − 2√3 + 1)/(3 − 1) = (4 − 2√3)/2 = 2 − √3 Especially you do not have to bother which sign (+ or −) is the correct one in equation [2.6].

@robertlynch7520

20 күн бұрын

@@unknownidentity2846 nicely done!! I'll keep that identity in mind.

This is awesome, many thanks, Sir! φ = 30°; ∎ABCD → AB = AE + BE = DF + CF = 40 = 2a; AD = EF = BC = a; sin⁡(DFE) = sin⁡(3φ) = 1 DFA = x = ? FCE = φ → CF = BE = a√3 → AE = DF = a(2 - √3) → √(2 - √3) = (√2/2)(√3 - 1) → AF = 2a√(2 - √3) → sin⁡(x) = a/2a√(2 - √3) = (√2/4)(√3 + 1) → cos⁡(x) = a(2 - √3)/2a√(2 - √3) = (√2/4)(√3 - 1) → sin⁡(2x) = 2sin⁡(x)cos⁡(x) = 1/2 = sin⁡(φ) = sin⁡(6φ - φ) = sin⁡(5φ) → sin⁡(x) = sin⁡(5φ/2) → x = 5φ/2

@PreMath

Ай бұрын

Thanks for sharing ❤️

x=75°

Other method ans won't be exactly 75 as yoy did sir.Thanks

@PreMath

Ай бұрын

Thanks for the feedback ❤️

arctan(20/(40-(20*tan(60))))

@PreMath

Ай бұрын

Thanks for sharing ❤️

At 4:18 , yes that is correct sir! Is not it? No, ...in uncontracted format. Contracted, one is an auxiliary verb (isn't) and the other a pronoun (it). Just sayin. ...I absolutely love the structure of language. 🙂

@PreMath

Ай бұрын

Excellent!😀 Thanks for the feedback ❤️

Tan 60=EB/20. EB=34,6. Hence AE=5,4. tan x=20/5,4. x=74,8. Short, isn t it?

@PreMath

Ай бұрын

Thanks for sharing ❤️

I did it with Trigonometry and got 74.89 which rounds off to 75 degrees.

@simpleman283

Ай бұрын

The answer is 75 exactly using trig. You did some rounding ro get 74.89.

@PreMath

Ай бұрын

Thanks for sharing ❤️

Tan x=20/(40-20sqrt(3))=1/(2-sqrt(3)), x=75.😅

@PreMath

Ай бұрын

Excellent! Thanks for sharing ❤️

Tan 75 =2+sqrt 3

@PreMath

Ай бұрын

Thanks for sharing ❤️

x=75°