Can you solve without using Trigonometry? | A Nice Geometry Problem
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- Опубліковано 4 чер 2024
- Can you solve without using Trigonometry? | A Nice Geometry Problem
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Rotate triangle CDE 90 degrees around point C. Point E will move to position E1. Obviously, ∡ECE1 = 90°. FE1 = x/2 + x/3 = 5x/6.
By the Pythagorean theorem: EF = sqrt(x^2/4 + 4x^2/9) = 5x/6. It follows that EFE1C is a deltoid. The diagonal CF is the bisector of the angle EСE1, so Θ = 90°/2 = 45°.
Very clever solution 👍
beautiful!
3rd method: Calculate areas of triangles AFE, BCF and DEC, then substract them from the full square area to get S[CEF]=9x²-3x²/2-3x²/2-9x²/4=15x²/4. By Pythagoras EC=√10x and FC=3√5x/2. Draw a height h from F to base EC. then h(EC)/2=15x²/4, h√10x/2=15x²/4 then h=3√10x/4. By Pythagoras calculate segment m at EC from C to the height, to get m=3√10x/4. Then m=h and obviously θ=45º.
good method
Rotate a copy of the Square by 90° Clockwise about C.
EF = FE', and ∠E'CF=90-θ
Since ∆ECF is Congruent to ∆E'CF by SSS, 90-θ = θ
Hence, θ = 45°
Nice solution, but you have to show that EF = FE', which is not too complicated:
EF is hypotenuse of △AEF, where AF = 3X/2, AE = 2X = 4X/2.This is a 3-4-5 triangle with EF = 5X/2
Since D' = B, then FBE' is a straight line, and FE' = FB + BE' = FB + D'E' = FB + DE = 3X/2 + X = 5X/2
And of course CF = CF and CE = CE' (since we are rotating about C, distances to C remain unchanged)
2:56 tan(alpha) ....You are using trigonometry, the video title said without trigonometry !
Join E toF. Find the areas of the 3triangles afe,fbc and edc and add.That gives (21/4)x^2. The area of ABCD=(9) x^2. Therefore the area of triangle efc equals (15/4)x^2. |ec|=root 10 times x and |fc|= (root 45 times x)/2. Area of triangle efc= ( root10/2)x mult. By (root 45/2)x mult by sine theta. Equate the two values for area of triangle efc and you get (root 450/4)by (x^2) by sine theta= (15/4)by(x^2). So sine theta = 1/(root2) giving theta = 45 degrees.
At 11:33, QM has been determined to be equal to X. Since PQ = 3X, that leaves PM = 2X. Construct ME. ΔPME and ΔQCM are congruent by side-angle-side (PE = QM = X,
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well, i have 6 squares, 3 x 2; and the lines to be extended to the corners, then, at a glance
Ιf I was a civil engineer 😊
Apply cosines formula in triangle EFC:
EF²=EC²+FC²-2EC⋅FC cosϑ =>
2⋅EC⋅FC⋅cosϑ = EC²+FC²- EF² (1)
EC²=10x² => EC=x√10
FC²=9x²+9x²/4=45x²/4 => FC=(3x√5)/2
EF²=4x²+9x²/4= 25x²/4
(1)=> 2⋅ x√10⋅(3x√5)/2 ⋅cosϑ= 10x²+45x²/4 - 25x²/4
=> 3x²⋅5√2 ⋅cosϑ=15x² => cosθ= 1/√2
=> cosθ=√2/2 => θ=π/4
Зачем такие сложности.Это задача решается очень элементарно. Берём треугольник CDE и вращаем вокруг точки С на 90° по часовой стрелке и получаем два одинаковых треугольника FEC и FCE'. У обеих треугольников все стороны равны а углы на вершине у одного альфа+бетта , а у другого тетта. А всё вместе 90°. Значит искомый угол 45°
Without loss of generality put AF=FB=3; AE=4; ED=2
Area of triangle EFC = 6*6 - 1/2 * (3*6+3*4+2*6)= 15
Let M be the perpendicular foot from E to FC then 15 =0.5 * FC * EM; Since FC= sqrt(9+36), EM= 30/sqrt45 = 10/sqrt 5 = 2*sqrt 5 ; EC = sqrt(4+36)= 2 * sqrt 10 = EM * sqrt 2
So Theta is 45 degree(since CM^2=EC^2-EM^2 => CM=EM if you absolutely want to avoid trig).
I was really not going to comment here but it is a simple problem that can be deduced almost by inspection.
Also note that in the second method Triangles PME is congruent to Triangle CQM so angle PME + angle CMQ=90; angle EMC=90 and EM=MC; Theta is 90/2= 45 degree.
This involves very little calculation whatsoever and not even Pythagorean.
*I want to prove that θ=45.* *It is enough to prove that θ is an acute angle of an isosceles right triangle* !!!!
EK⊥FC construction Area (EFC)=(ABCD)-(AEF)-(BFC)-(EDC)
= 9x²-(3x²)/2-(9x²)/4-(3x²)/2
(EFC)=(15x²)/4 (1)
In orthogonal triangle apply Pythagoras theorem :
FC=(3√5 x)/2
Now area (EFC)=(EK⋅FC)/2 =>
(EFC)=EK (3√5 x)/4 (2)
(1),(2)=> EK (3√5 x)/4 = (15x²)/4
=> *EK= x√5* (3)
Apply Pythagoras theorem in ΔEKC =>
KC²=EC²-EK²
=10x²-5x²
*KC=x√5* (4)
(3),(4) => EK=KC => right triangle EKC
is isosceles = > θ=45° finish
I think that I can do that WITHOUT trigonometry. And I think that this looks easier than expected. All that you have to do is to make a chord that intersects with on of the other chords, subtend right angles, then use the angle bisector extension of the circle theorem to use AA similarity of two right angled triangles, then make second and chord that forms two CONGRUENT triangles that are congruent simply by the definition of what a supplementary angle is. The congruency is justified from applying the Pythagoras theorem twice AFTER you determine the sides of the two similar triangles. I think that that is the best summary that can think of.
Also the first chord has to have the same sides in order for right angles to be subtended. And you have to use the definition of a supplementary angle once you establish equal sides and angles. And this is done in order to check the triangle that subtends theta is equilateral equiangular or a right scalene triangle. And because the right scalene pair has congruent angles, you need the definition of a supplementary angle, correct?
With trigonometry it would be very easy
asnwer= 45
Wtf only now you are saying ABCD is a square. Why did you not state it originally asso
Как видно из решений, второй метод дополнительных графических построений, даёт быстрый 100% результат решения, так как он наглядный и более понятный для восприятия. сложнее найти точные значения прилежащих углов a и в, так как тригонометрический калькулятор на даст точные значения этих углов по тангенсам, синусам, и прочей дребедени. Если при расчётах космических полётов, ошибиться на тысячную долю угла запуска, относительно плоскости эклиптики, то без коррекции орбиты полёта. можно улететь не туда!
🤔😅😂🤣