Find the angle X | A Nice Geometry Problem | 2 Different Methods to Solve 

Add a line CD to make two isosceles triangles in which we see that ACcos 2x=AD/2, so cos 2x=sqrt(3)/2=cos 30, so 2x=30, x=30/2=15.😊

At 3min, drop a perp from C to AD. You have 2 triangles where cos 2x = rt3/2, so 2x =30deg.

The law of sines: sin(pi -3x)/(1 + sqrt(3) ) = sin (x), is one equation in one unknown. sin(pi/12) (1 + sqrt(3) ) = 1/sqrt(2) = sin ( 3 pi/4). One doesn't need any additional constructions.

ad=√3 . drop cp perpendicular to ad . ap=√3/2. Cos2x=√3/2 means x=15 degrees

Thanks for your interesting problem. Here is the way I solved it using pure trigonometry, of course without looking at your solution. I hope you like it. Greetings. Problem recall triangle abc has got following properties angle a = a = 2x angle b = b = x angle c = c = unknown side AC = 1 side AB = sqrt(3)+1 Let's recall that in a triangle, the angle sum is PI = a+b+c so angle c=PI-(a+b)=PI-(2x+x)=PI-3x c=PI-3x Let's write sinus law sina/BC=sinb/AC=sinc/AB=sin2x/BC So applying for triangle ABC we have sinx/1=sin(PI-3x)/(sqrt(3)+1) as sin(PI-y)=siny we have sin2x/BC=sinx=sin3x/(sqrt(3)+1) (1) sinx=sin3x/(sqrt(3)+1) Let's calculate sin(3x) sin(3x)=sin(2x+x)=sin(2x)cosx+sinxcos(2x) sin(3x)=2sinx.cosx.cosx+sinx.(1-2(sinx)^2) sin(3x)=2sinx.(cosx)^2+sinx.(1-2(sinx)^2) sin(3x)=2sinx.(1-(sinx)^2)+sinx.(1-2(sinx)^2) (2) sin(3x)=3sinx-4(sinx)^3=sinx(3-4(sinx)^2) Equalities (1) and (2) give us sinx=sin3x/(sqrt(3)+1)=sinx(3-4(sinx)^2)/(sqrt(3)+1) (3) sinx=sinx(3-4(sinx)^2)/(sqrt(3)+1) so we have sinx=0 or [sinx0 and sinx=sinx(3-4(sinx)^2)/(sqrt(3)+1)] So there is three cases sinx=0 (x=0 or x=PI) and sinx0 case 1: If we take angle b=x=0, then angle a=2x=0 The triangle solution is a flat triangle: -with points A, B and C alined with C between A and B -with following lengths AB=sqrt(3)+1, AC=1 and BC=sqrt(3)=AB-AC Case 1 is a solution case 2: If we take angle b=x=PI, then angle a=2x=2PI=0 [mod 2PI] The triangle solution is a flat triangle: -with points A, B and C alined with B between A and C -this case is not possible according to the hypothesis lengths as AC=1

Solution using pure Geometry - Let E = midpoint of AB, let CD is perpendicular to AB, let F = midpoint of CA. Hence, F is circumcenter of ∆CAD. Hence, CF = AF = FD = 1/2. ∆FAD is isosceles. Hence, measure of angle FDA = 2x. FE is || CB ( by midpoint theorem). Hence, measure of angle FEA = x = measure of angle CBA. But measure of angle FDA = 2x. Hence, by external angle theorem, measure of angle DGE = x. Hence, ∆FDE is isosceles. Hence, FD = DE = 1/2. Also, AE = (√3 + 1)/2 Hence, AD = (√3 + 1)/2 - 1/2 = (√3/2). Hence, ∆CAD is 30-60-90 ∆. Hence, A = 2x = 30. Hence, x = 15.

√3 + 1 = AB/ AC = sin (3x) / sin ( x) Hereby cos(2x) + 2 cos^2 ( x) = √3 + 1 2 cos (2 x) + 1 = √3 + 1 cos (2 x) = √3/2 x = arc cos ( √3 / 2) / 2 = 15°

1st method: Draw CD, where ∠ADC = 2x. As ∠CAD = ∠ADC, ∆DCA is an isosceles vtrisngle and CD = AC = 1. ∠ADC is an external angle to ∆CDB at D, so ∠ADC = ∠DBC + ∠BCD ∠ADC = ∠DBC + ∠BCD 2x = x + ∠BCD ∠BCD = 2x - x = x As ∠BCD = ∠DBC = x, ∆CDB is an isosceles triangle and DB = CD = 1. As DB = 1 and AB = √3+1, AD = √3. Let ∠DCA = α. cos(α) = (AC²+CD²-AD²)/2AC•CD cos(α) = (1²+1²-(√3)²)/2(1)(1) cos(α) = (2-3)/2 = -1/2 α = cos⁻¹(-1/2) = 120° α + 2x + 2x = 180° 120° + 4x = 180° 4x = 180° - 120° = 60° x = 60°/4 = 15° 2nd method: ∠A = 2x and ∠B = x, so ∠C = 180°-3x. (√3+1)/sin(180°-3x) = 1/sin(x) = BC/sin(2x) (√3+1)/sin(180°-3x) = 1/sin(x) sin(180°-3x) = (√3+1)sin(x) sin(3x) = (√3+1)sin(x) 3sin(x) - 4sin³(x) = (√3+1)sin(x) 3 - 4sin²(x) = √3 + 1 4sin²(x) = 3 - √3 - 1 = 2 - √3 sin²(x) = (2-√3)/4 = (4-2√3)/8 sin(x) = √((4-2√3)/8) sin(x) = √(3-2√3+1)/2√2 sin(x) = √(√3-1)²/2√2 sin(x) = (√3-1)/2√2 = √3/2√2 - 1/2√2 sin(x) = (√3/2)(1/√2) - (1/2)(1/√2) sin(x) = sin(60°)cos(45°) - cos(60°)sin(45°) sin(x) = sin(60°-45°) = sin(15°) x = 15°

😊

15 degrees. I did it in my head.

Con carnot e il teorema dei seni risulta (cosx)^2=(7+4√3)/(8+4√3)...x=15

شکل غلط است