Can you find angle X in the square ABCD? | Geometry and Trigonometry skills explained | (Fast!)

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Can you find angle X in the square ABCD? | Geometry and Trigonometry skills explained | (Fast!)

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[marioalb]

tan α = y / 3y = 1/3
α = 18, 435°

tan β = 1, 5y / 3y = 1/2
β= 26, 565°

Angle X = 90° - α - β
Angle X = 45° ( Solved. √ )

Is not necessary to use the trigonometric identity !!, just add angles

[tellerhwang]

用全等也可解

E'B=EB, 角FBE'=角CBE+角FBA
且F, A, E共線(角C+角A=180)
E'F=5y/2, 角FBE'=角CBE+角FBA
2.連接F, E, 由畢氏定理知EF=5y/2
→EF=E'F, EB=E'B, FB=FB
△EFB全等△E'FB(SSS)
→角x=角FBE'→x=45

[johnbrennan]

Join F to E.Find the lengths of the three sides of the triangle FEB in terms of y. Then use the cosine rule. Cosine x = 1/sqroot 2.So x= 45 degrees.

[partyprop]

can also be solved by congruence

1. With B as the fixed point, rotate △ECB until CB overlaps with AB, then △E'AB

Congruent △ ECB

E'B=EB, FBE'=CBE+FBA

BFA, E(C+A=180)

E'F=5y/2, FBE'=CBE+FBA

2.FEEF=5y/2

→EF=E'F, EB=E'B, FB=FB

A EFBA E'FB(SSS)

1

x=FBE-x=45

[williamwingo]

x = 90 --- arctan(1/2) -- arctan(1/3).= 90 -- 26.56 -- 18.43 = 45.01. Do it in your head, up to the point of calculating the arctangents: 🤠

[misterenter-izrz]

Without lost of generality, let y=2, then BE^2=36+9=45, EF^2=16+9=25, BF=36+4=40, then cos x=(45+40-25)/2root 45 root 40=5/ root 50=1/root2, therefore x=45.😮

[wackojacko]

Geometry and Trigonometry...powerful tools indeed! 🙂

[Q_from_Star_Trek]

I calculated with Pitagora's theorem BE =3sqrt5/2*y, BF=sqrt10*y, EF=5/2*y; Law of cosines theorem EF square = BE square + BF Square -2*BE*BF*cos (x)>>>cos(x) =1/sqrt (2)>>>x =45 degrees.

[harikatragadda]

A geometric solution would be to rotate a copy of the square by
90° anti-clockwise about B.
Now, FE = FE' = 5y/2
Implying
∆FBE' is Congruent to ∆FBE
Hence, ∠FBE' = FBE = α + β = X
Since X + α + β = 90°
X = 45°
Image here

[samimarzou]

Let a be side of square ABCD
Area EBF = Area square ABCD - Area (ECB+AFB+DEF) = a^2 - ((a2/2 × a)/2 + (a × a/3)/2 + ((a/2)×(2a/3))/2) = a^2 - (a^2/4 + a^2/6 + a^2/6) = 5a^2/12 (i)
Also area EBF = 1/2 EB × BF × sinx
EB= a sr(5)/2 and BF= a sr(10)/3 as determined by Pythagoras theorem, therefore
EBF= 1/2 ×( a^2/6) × sr (50) × sin x = 5a^2/12 × sr(2)× sinx (ii)

Equate (i) and (ii)
(5a^2/12) = (5a^2/12)× sr(2) × sin x >
sin x = 1/sr(2) = sr(2)/2, therefore x = 45

[bekaluu]

We can also find alpha and beta separately using arctan and subtract the sum of the two to arrive at the same solution of 45 degrees. Thanks

[SaadmaanMahmidShoshi]

I used the inverse tangent function to calculate the approximate values of angles alpha and beta, then simply plugged the values into alpha + beta + x = 90. Still gave me the right answer, which is x = 45 degrees. 😄

[marioalb]

tan α = y / 3y = 1/3
tan β = 1, 5y / 3y = 1/2

Angle X = 90° - α - β
X = 90° - atan(1/3) - atan(1/2)
X = 45° ( Solved. √ )

[jorowh]

Define this square as a unit square with B as the origin (0, 0). Extend BF to point F' (-3, 1) . Extend BE to point E' (-1, 2). Now F'E'B form a 45, 45, 90 triangle, one angle of which is our required angle x.

[DraCat]

We draw the line EF. if we take y =2, then the side of the square is 6, we subtract the areas of the outer triangles from the square area. The area of the triangle EFB is equal to 15. through the Pythagorean theorem, we find BF and BE, they are equal to 2√10 and 3√5. 1/2*(2√10*3√5* Sinx)=15. SinX=30÷6√50, sinX=√2/2. X=45°.
I know that the method is not universal. Perhaps the text is written illiterate. I write through a translator. Hello from Russia

[vkr]

Ģeometrisks risinājums: BE pagarinam līdz krusto AD apzīmējam X, XDE=BCE, XF=5y. BCE rotejam ap B līdz C=A, E1F=5y/2. BE1/BX=E1F/FX tātad BF bisektrise, .x=45 grādi.

[devondevon]

x=45 answer
since x + the two smallest angles ( of the two triangles) = 90
Assume the side of the square is 18 ( I use 18 since it is divisible by 2 and 3 ( or 6) to avoid using fractions since the square has one side divided by 2 and the other side divided by 3), then the triangle to the right has sides 18 and 9. The third side is 20.12 (Pythagorean Theorem)
The triangle below sides are 18 and 6, and the third side is 18.97.
Let the smallest angle of the triangle to the right = p, and the smallest angle of the
triangle below = n, then x= 90 - (n+p).
To calculate n, 18, 9, and 90 degrees will be used, and the Law of Cosine since there are two sides and angles.
This equals 26.565 degrees.
To calculate p, 18, 6, and 90 degrees will be used, and the Law of Cosine since there
are two sides and an angle.
This equal 18.435
18.435 + 26.565 =45
90-45 = 45 Answer

[MrPaulc]

I went simpler: 90 - arctan(1/3) - arctan(1/2) = 45 degrees. As in 90 - arctan (y/(3y)) - arctan(y/(2y))

[EngNALrashed]

Different solution but same results 👏🏻👏🏻👏🏻

[murdock]

Nice!

∎ABCD → AB = BC = CD = AD = a; DE = CE = a/2; DF = 2AF = 2a/3
fast lane: tan⁡(δ) = AF/AB = 1/3; tan⁡(φ) = CE/BC = 1/2 →
tan⁡(δ + φ) = (tan⁡(δ) + tan⁡(φ))/(1 - tan⁡(δ)tan⁡(φ)) = 1 → δ + φ = 45° = x
or:
∆ABF → ABF = δ → BF = a√10/3 → sin⁡(δ) = √10/10 → cos⁡(δ) = √(1 - sin^2(δ)) = 3√10/10
∆BCE → EBC = φ → BE = a√5/2 → sin⁡(φ) = √5/5 → cos⁡(φ) = √(1 - sin^2(φ)) = 2√5/5 →
sin⁡(δ + φ) = sin⁡(δ)cos⁡(φ) + sin⁡(φ)cos⁡(δ) = √2/2 = cos⁡(δ + φ) →
δ + φ = 45° → x = 90° - (δ + φ) = 45°
or:
EF = √((a/2)^2 + (2a/3)^2) = 5a/6 →
(25/36)a^2 = (10/9)a^2 + (5/4)a^2 - 2(a√10/3)(a√5/2)cos⁡(x) = (1/36)a^2(85 - 60(√2)cos⁡(x)) →
25 = 85 - 60(√2)cos⁡(x) → cos⁡(x) = sin⁡(x) = √2/2 → x = 45°