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
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)
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
@@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
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 ❤
Both solutions are nice. The second one is amazing. Congrats
Thank you 🙂
In The 3rd method you construct a triangle CBE that has an angle 2 theta. ¿When do you conclude that EC contains D ?
You make a symmetry respect BH orthogonal to AC and you will obtained a vertex E that fulfills the conditions.
2√2cos(theta) =[tan(theta) -1]/tan(theta)
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 < 2θ < 1
*sin(2θ) = 1/2 → 2θ = 30° → θ = 15°*
Is BD perpendicular to AC?
no
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)
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
@@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
this was painful to watch