Justify your answer | Find the angle X | #math #maths | #geometry

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Justify your answer | Find the angle X | #math #maths | #geometry

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[marcelowanderleycorreia]

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

[kaziummeruman]

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

[prossvay]

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.

[dirklutz]

You made it so easy again. Great idea!

[quigonkenny]

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°

[yalchingedikgedik]

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

[marcgriselhubert]

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.

[phungpham]

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😅

[jimlocke]

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 <AFE, opposite the short side, is 15°. <AFD and <AFE are complementary, so x = <AFD = 75°, as PreMath also found.

The 15°-75°-90° right triangle appears often enough in geometry problems that I recommend that we all memorize its properties. Long side = (2 + √3) times the short side. Expressed differently, ratio short side : long side :hypotenuse = (√3 - 1):(√3 + 1) : 2√2, if the hypotenuse is the base, the height is 1/4 of the hypotenuse and area = 1/8 of the hypotenuse squared.

[waheisel]

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!

[MrPaulc]

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.

[LuisdeBritoCamacho]

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.

[santiagoarosam]

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

[simpleman]

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

[unknownidentity]

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

[sergeyvinns]

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

[murdock]

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

[wackojacko]

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. 🙂

[rajendraameta]

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

[AndreasPfizenmaier-yw]

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