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°
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
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.
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.
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 ✓ ≡=-⋅
@@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].
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!
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. 🙂
Very good solution!! We have to think outside the box.
Yes, exactly
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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.
Excellent!
You are very welcome!
Thanks for sharing ❤️
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°
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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.
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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
Excellent!
Thanks for sharing ❤️
You made it so easy again. Great idea!
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.
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Thanks Sir
That’s nice
With glades
❤❤❤❤
Thank you!
You're welcome!❤️
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😅
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...
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Thank you for the proof
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Thanks ❤️
Why we use alpha in this geometry?? Plz answer😥😥
x=75°
Yes, got it
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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.
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
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 ✓ ≡=-⋅
Awesome!
Very much appreciated!
Thanks for sharing ❤️
@@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].
@@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
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tgx=20/(40-20tg60)=1/(2-√3)=2+√3
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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!
You are very welcome!
Thanks for sharing ❤️
FB=40, angle ABF= 30, angle FAB=angle X= (180-30)/2=75!
Excellent!
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Tg (X)=20/(40-20√3)=2+√3 ; X=75°
Gracias y saludos.
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I did not catch both legs being 40.
I should have known you would have a sneaky way of doing it.
👍
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Tan 60=EB/20. EB=34,6. Hence AE=5,4. tan x=20/5,4. x=74,8. Short, isn t it?
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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. 🙂
Excellent!😀
Thanks for the feedback ❤️
arctan(20/(40-(20*tan(60))))
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Wow so much easier than invtan (2+√3)
Glad to hear that!
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Tan x=20/(40-20sqrt(3))=1/(2-sqrt(3)), x=75.😅
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Other method ans won't be exactly 75 as yoy did sir.Thanks
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Wonderful. Continuously watching this channel for a week is guaranteed to increase IQ by 1 quotient
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Glad to hear that!
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I did it with Trigonometry and got 74.89 which rounds off to 75 degrees.
The answer is 75 exactly using trig.
You did some rounding ro get 74.89.
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Tan 75 =2+sqrt 3
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x=75°