My approach Measures of angles in triangle ADB Angle DBA = 3theta Angle DAB = 90-2theta Angle ADB = 90 - theta From sine rule in ADB (x-y)/sin(3theta) = y/sin(90-theta) (x-y)/y = sin(3theta)/cos(theta) Express sin(3theta)/cos(theta) in terms of tan(theta) From triangle ABC we know that tan(2theta) = y/x x/y - 1 = sin(3theta)/cos(theta) x/y - 1 = 1/tan(2theta) - 1 After comparing these two results we will get quartic equation easy to solve but only tan(theta) = 2 - sqrt(3) will be valid solution so theta = 15

Very nice solution ❤

In △ABC, ∠C = 2θ, ∠B = 90 → ∠A = 90−2θ In △ABC, tan(2θ) = y/x → *x/y* = cot(2θ) = *cos(2θ)/sin(2θ)* In △ABD, ∠A = 90−2θ, ∠B = 3θ → ∠D = 90−θ Using law of sines in △ABD we get: AD/sin(∠B) = AB/sin(∠D) AD/AB = sin(∠B)/sin(∠D) (x−y)/y = sin(3θ)/sin(90−θ) x/y − 1 = sin(3θ)/cosθ cos(2θ)/sin(2θ) − 1 = sin(3θ)/cosθ cos(2θ) − sin(2θ) = sin(3θ) sin(2θ)/cosθ = sin(3θ) * 2 sinθ cosθ / cosθ cos(2θ) − sin(2θ) = 2 sin(3θ) cosθ Using product-to-sum identity, we get: cos(2θ) − sin(2θ) = cos(2θ) − cos(4θ) sin(2θ) = cos(4θ) sin(2θ) = 1 − 2 sin²(2θ) 2 sin²(2θ) + sin(2θ) − 1 = 0 (sin(2θ) + 1) (sin(2θ) − 1) = 0 2θ = −1 or 1/2 Since 2θ is an acute angle (it is a non-right angle of right triangle), then 0 *sin(2θ) = 1/2 → 2θ = 30° → θ = 15°*

2√2cos(theta) =[tan(theta) -1]/tan(theta)

In The 3rd method you construct a triangle CBE that has an angle 2 theta. ¿When do you conclude that EC contains D ?

@user-yl6zt7vq4t

3 ай бұрын

You make a symmetry respect BH orthogonal to AC and you will obtained a vertex E that fulfills the conditions.

I love watching your videos💛

Dai teorema dei seni risulta (sin3θ/cosθ)+1=ctg2θ..dopo le semplificazioni(θ=-45non è accettabile)risulta(tgθ)^2-4tgθ+1=0..tgθ=2+√3,θ=75(??)...tgθ=2-√3,θ=15(ok)

@holyshit922

3 ай бұрын

I used the same approach but i desribed it more exactly and took me longer Although we used the same method i solved it independently Our approach is easier than presented on the video but some people may like his solutions In fact his first method is not so difficult to get His second method is focused on isosceles triangles and is more difficult then first method because we do not see at the first sight why we should look for isosceles triangles

@giuseppemalaguti435

3 ай бұрын

​​@@holyshit922x/cosT=t/sin2T... y/cosT=t/cos2T.. Divido le 2 equazioni x/y=ctg2T.. Inoltre y/cosT=(x-y) /sin3T.( x-y) /y=sin3T/cosT=x/y-1...quindi ctg2T=1+sin3T/cosT..this is the final equation.. My approach is, perhaps, not very simple.. But for me, is natural

Thita =15 degree, may be

Is BD perpendicular to AC?

@georgesdellopoulos5808

3 ай бұрын

no

this was painful to watch