Your construction inspired me to find a second method: In ∆BPQ, sin theta=x/BP In ∆BAC, sin theta=y/AB So, x/BP=y/(x+BP) => x.(x+BP)=y.BP But, x.(x+BP)=y.y from semicircle Hence, BP=y => y^2=x^2+x.y etc.
This is a math olympiad problem that shows how useful a golden angle really is!!! The angle theta is arcsin[(sqrt(5)-1)/2]. Also I wish that there was a mnemonic device for understanding HL congruence involving beta and the right angle. No lie. I shall use this for practice!!!
2:30 ∆PQB is Congruent to ∆APC, hence PC =PB If BQ=a and QC=b, then by the Similarity of ∆PQB and ∆CQP PQ/a=a/PQ PQ=√(ab) In ∆PQC, a²=b²+(√(ab))² (b/a)²+(b/a)-1=0 b/a= ½(√5 -1) Sinθ=QC/PC=b/a=½(√5 -1) θ≈38.17°
PQ=a...r=raggio..2rcosθ=a/sinθ..r=(a/2)/sinθcosθ...(a/sinθ):a=((a/sinθ)+a):2rtgθ...semplifico la a,risulta...1/sinθ=(1+1/sinθ)/(tgθ/sinθcosθ)..1/sinθ=(cosθ)^2(1+1/sinθ)... calcoli (sinθ)^2+sinθ-1=0..sinθ=(-1+√5)/2...θ=arcsinφ=38,17
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Your construction inspired me to find a second method:
In ∆BPQ, sin theta=x/BP
In ∆BAC, sin theta=y/AB
So, x/BP=y/(x+BP)
=> x.(x+BP)=y.BP
But, x.(x+BP)=y.y from semicircle
Hence, BP=y => y^2=x^2+x.y etc.
This is a math olympiad problem that shows how useful a golden angle really is!!! The angle theta is arcsin[(sqrt(5)-1)/2]. Also I wish that there was a mnemonic device for understanding HL congruence involving beta and the right angle. No lie. I shall use this for practice!!!
And how did you *use* the Golden Angle_ to solve the problem?
@@hanswust6972Oh I was only commenting that this is an angle that theta is simply the arcsin of phi or phi inverse. Which is (sqrt(5)-1)/2.
2:30 ∆PQB is Congruent to ∆APC, hence PC =PB
If BQ=a and QC=b, then by the Similarity of ∆PQB and ∆CQP
PQ/a=a/PQ
PQ=√(ab)
In ∆PQC, a²=b²+(√(ab))²
(b/a)²+(b/a)-1=0
b/a= ½(√5 -1)
Sinθ=QC/PC=b/a=½(√5 -1)
θ≈38.17°
Construct PC. Drop a perpendicular from P to AC and label the intersection as point R. PR and BC are parallel.
Excellent method.
PQ/PC = cosθ=PA/PC = tanθ
(cosθ)^2 =sin θ
1-(sinθ)^2 = sinθ
So :
(sinθ)^2+sinθ-1 = 0 ….. etc
No need for the secant theorem
PQ=a...r=raggio..2rcosθ=a/sinθ..r=(a/2)/sinθcosθ...(a/sinθ):a=((a/sinθ)+a):2rtgθ...semplifico la a,risulta...1/sinθ=(1+1/sinθ)/(tgθ/sinθcosθ)..1/sinθ=(cosθ)^2(1+1/sinθ)... calcoli (sinθ)^2+sinθ-1=0..sinθ=(-1+√5)/2...θ=arcsinφ=38,17
tanθ = (θ.cosθ)/ θ
tan θ = cos θ
sin θ = cos² θ = 1 - sin²θ
sin² θ - sin θ -1 = 0
sin θ = 0,61803
θ = 38,17° ( Solved √ )
Calculators allowed!
Çok eski bir soru bu.
Çözüm şu ana kadar bilinmiyordu.
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@@hanswust6972 niye bilinmesin ki?
180°/180°ABC=1ABC (ABC ➖ 1ABC+1).