I did it the second way, but the first way is much more elegant. Always prefer a geometric solution over a trigonometric one. Just more beautiful.

This problem is simple if we use an adapted orthonormal. We choose center D and frst axis (DC). The problem is independant of the length DC = BD, we choose this length equal to 1 for example. So the equation of (CA) is y = -tan(30°).(x -1) or y = (-sqrt(3)/3).x + sqrt(3)/3. The equation of (DA) is y = -x. At point A, at the intesection, we have: (3 -sqrt(3)).x = sqrt(3) so x = (1 +sqrt(3))/2 when simplified. So we have A((1+sqrt(3))/2; -(1+sqrt(3))/2) As B(-1;0) then VectorBD((3+sqrt(3))/2; -(1+sqrt(3))/2) (form (m;n)), m/n = -1/sqrt(3) = -sqrt(3)/3 is equal to tan(theta), so theta = 120°

Being AEC a 30,60,90 degree right triangle if AE = X => EC = X√ 3 Being AED a right isosceles triangle AE = ED = x setting EB = y we can write the following identity: x - y = x√ 3 - x (because BD = CD) y = 2x - x√ 3 Considering triangle AEB we can write: tan (180 - theta) = x/(2x - x√ 3) = 2 + √ 3 that means that 180 - theta = 75° theta = 180 - 75 = 105

Just one question, could the adapted orthonormal be applied to the last geometry problem?

Application of sine rule can as well solve the problem. Sin(30)/sin(15)=sin(theta)/sin(135-theta) Then using trig. identity for sin(135-theta). Theta=-75 which is equivalent of 105 following anticlockwise measure. Please confirm

Excellent solutions🎉🎉🎉

My method: Let H be the height of triangle ABC with respect to A (extending the BC line and tracing the height from point A). Let E be the intersection point between H and BC line's extension. And X = BD = DC DE = H (equilateral triangle) BE = H-X Tg30° = H/(H+X) Doing some algebra, we get the relation: X= H(root3 - 1) Lets see the tangent of angle ABE = 180°- tetha Tg (180-tetha) = H/(H-X) Lets substitute X Tg (180-theta) = H/(H - H(root3-1)) Tg (180-tetha) = 1/(2-root3) Simplifying Tg (180-tetha) = 2+ root3 I know that tg 75° = 2+root3 So, Tg(180-tetha)= tg 75° 180-tetha=75 Tetha= 105°

@gabri41200

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Oh, now i saw the video, and this is the second method 😅

second method

ctgθ=(√2sin15/sin30)-1..θ=-75..θ=105

I did second method before checking the solution, but first method is much more elegant

tgθ=-2-√3...θ=-75...θ=105

Easy、but good question🙂

As ∠BDA is the external angle to ∆ADC at D: ∠BDA = ∠DCA + ∠CAD 45° = 30° + ∠CAD ∠CAD = 45° - 30° = 15° As BD is a straight line, ∠ADC = 180°-45° = 135°. Draw DE, such that E is the point on CA where ∠CED = 30°. As ∠CED = ∠DCE = 30°, ∆EDC is an isosceles triangle, DC = DE, and ∠EDC = 180°-(30°+30°) = 120°. As ∠ADC = 135°, ∠ADE = 135°-120° = 15°, so ∠ADE = ∠EAD, ∆DEA is an isosceles triangle, ∠DEA = 180°-(15°+15°) = 150°, and DE = EA. As BD = DE, then ∆BDE is an isosceles triangle. As ∠ADE = 15° and ∠BDA = 45°, ∠BDE = 60°, so ∠DEB = ∠EBD = 60° as well, and ∆BDE is an equilateral triangle, and EB = BD = DE. As BE = EA, ∆BEA is an isosceles triangle. As ∠DEA = 150° and ∠DEB = 60°, ∠BEA = 150°-60° = 90°, so ∠ABE = ∠EAB = 45°. ∠ABD = ∠EBD + ∠ABE θ = 60° + 45° = 105°

45-30=15+90=105÷360×314.159268

120

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