AC=6/sin t, EC=3/sin t, EC/2=tan t, 3/2 1/sin t =tan t, 3/2=tan t sin t= sin^2/cos t=(1-cos^2 t)/cos t, 3cos t=2-2cos^2 t, 2cos^2 t+3cos t-2=0, (2cos t-1)(cos t+2)=0, so t=60.😊

|DC|=sqroot (a^2 + 4).Triangles ABC and DCE are similar. So 6/2a = a/ sqroot (a^2 + 4) .Then 3/a = a/ sqroot (a^2 + 4). That gives a^2=3(sqroot a^2 + 4). Square both sides. So a^4 = 9( a^2 + 4) .Therefore a^4 - 9a^2 - 36= 0. That means (a^2 - 12)(a^2 + 3)= 0. So a^2-12= 0. That gives a= sqroot 12 = 2 root 3. In triangle DCE, tan theta = 2 root3/ 2= sqroot 3. This implies theta= 60 degrees.

EDC and ABC are similar (by angle angle congruency theorem) So 2/EC=AB/6 Let AB= a and EC=b 12= ab b=12/a ABC is a right triangle So by pythagorean theorem AC^2=a^2+36 4b^2=a^2+36 4(144/a^2)=a^2+36 Let a^2= c 4(144/c)=c+36 Multiply both sides by c 576=c^2+36c 0=c^2+36c-576 0=c^2-12c+48c-576 0=c(c-12)+48(c-12) 0=(c+48)(c-12) c+48=0 or c-12=0 c=-48 or c=12 a^2=-48 or a^2=12 (Since a>0) Then a^2=(2*sqrt3)^2 So a=2*sqrt 3 tan @= 6/2*sqrt 3 = 6*sqrt 3/2*3 =(6*sqrt 3)/6 =sqrt 3 @= arctan (tan @) =arctan (sqrt 3) =60° (@

Say AE=EC=x Angle EDC is theta as well. In triangle ABC sin(theta)= 6/2x In triangle DEC tan(theta)=x/2 Eliminating x in the two equations you will get a trigonometric quadratic equation in terms of cos(theta). In fact the coefficients are the same as your equation. The solution would be theta being 60 degrees. No construction lines!!

Если в заданном треугольнике ABC отрезок DE равен (⅓) от катета BC, то в нём: ∠θ=60°, ∠α=30°, а сам △ABC - это половина равностороннего треугольника. Прежде всего проверим этот случай: DE:BC = 2:6 = ⅓. Всё так и есть. Значит ∠θ=60°.

Nice video and presentation. This is not just geometry quiz but algebra intense problem. None I watched is solely geometric.

E = Midpoint AC = center of circumference of △ABC (Thales). Draw a perpendicular from E to BC = point F. F = midpoint of AC. Call x = EF and y = DF, we have x²+y²=2²=4 AND similar triangles △DEF and △CEF, hence x/3 = y/x → x²=3y → x² = 4-y² = 3y, with y>0 we get 4-y=3 → y=1. Then x = √(2²-1²) = √3 and AB = 2√3, hypotenuse △DEF = 2 = 2y, therefore △DEF is a 30-60-90° triangle, which is similar to △ABC (and △CEF), hence ∠BAC = 60° and ∠BCA = 30°.

60 My approach Let AC = 2a, then AE = a and CE=a Triangle ABC and triangle CDE are similar Hence 2/a= AB/6 a AB =12 AB = 12/a (12/a)^2 + 36 = 4a^2 144/a^2 + 36 = 4a^2 Pythagorean Theorem 144 + 36a^2= 4 a^4 36 + 9a^2 = a^4 a^4 - 9a^2 - 36=0 Let a^2 = n n^2 - 9n-36 (n-12)( n+3) =0 n = 12 and n=-3 Hence a^2 = 12 a= sqrt 12 Hence AC = 2 sqrt 12 since AC =2a Since BC = C We can find AB = 3.4641 or sqrt 12 Notice that 2 sqrt 12 is twice 3.4641 or sqrt 12 To find out if this is a 30-60-90 right triangle multiply sqrt 12 by sqrt 3 = sqrt 36 = 6 , So this is a 30-60-90 right triangle . Answer = 60 degrees

It can be solved with Analytic Geometry ,with vectors . If you estmate scalar product of vectors ΑΒ and ΑΕ and their vectors norm , you will find cosθ=1/2 , so θ=60°

My solution: Let AE = CE = a, CD = x, where a, x > 0 △ABC ~ △DEC (by AA) as follows: ∠ACB = ∠DCE (same angle) ∠ABC = ∠DEC = 90° (by definition) From similar triangles: AC/DC = BC/EC 2a/x = 6/a → a² = 3x (**) Using Pythagorean Theorem in △DEC, we get DE² + CE² = CD² 2² + a² = c² → Replace a² with 3c [from (**)] 4 + 3c = c² c² − 3c − 4 = 0 (c − 4) (c + 1) = 0 c = 4 (since c > 0) a² = 3c = 12 → a = √12 = 2√3 Now we can find θ using ratio of sides of △ABC sin θ = BC/AC = 6/(2a) = 6/(4√3) = √3/2 θ = sin⁻¹(√3/2) = 60°

Brilliant!

What in the beginning seemed to be an easy problem came out to be difficult but thanks to your detailed explanation it was easily understood Thank you

Another method: △ABC ~ △ DEC Thus AB/DE=BC/EC i.e., AB/2=6/EC -- (i) Let's suppose AB=x. From (i), we get: EC=12/x AC=2*EC => AC=24/x Applying Pythagoras' theorem to △ABC, we get: x²+6²=(24/x)² x²+36=576/x² Muliplying throughout by x², we get: x⁴+36x²-576=0 i.e., (x²+48)(x²-12)=0 Because x²>0, we discard -48. Hence x²=12 or x=2√3 Thus AB=2√3, BC=6 => tan(θ)=BC/AB=6/(2√3)=√3=> θ=60° (because tan60°=√3)

Reflecting ∆ABD about the line AD forms ∆AED. Hence, AB = AE, Cosθ= ½, θ= 60°

Or you can apply sine rule to the large and small triangles then eliminate x. Same quadratic equation results.

ABED is cyclic using power of point 6√(a^2+4)=2a^2 => a=2rt3 draw EM parallel to BC Em=3 by midpoint theorem then sinA=rt3/2

60 degrees as this is a 30-60-90 right triangle

شكرا لكم على المجهودات يمكن استعمال جيب thêta..... هذا الجيب هو جذر ثلاثة على اثنين اي thêta هي ستون درجة

BE=EA=EC so ABE is an equilateral triangle so alpha is 60 degrees. Solved it less than 60 seconds

BC is not needed!

60 degrees

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