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Question Number 184398 by HeferH last updated on 06/Jan/23

Answered by som(math1967) last updated on 06/Jan/23

 ((BD)/(BC))=((sin2φ)/(sin(180−3φ)))=((sin2φ)/(sin3φ))  ((AD)/(BD))=((sin90)/(sin(90−3φ)))=(1/(cos3φ))   BD=ADcos3φ   ((ADcos3φ)/(AD))=((sin2φ)/(sin3φ))   [∵BC=AD]  sin3φcos3φ=sin2φ  2sin3φcos3φ=2sin2φ  sin6φ=2sin2φ  3sin2φ−4sin^3 2φ=2sin2φ  ⇒3−4sin^2 2φ=2   [φ≠0]   sin2φ=(1/2)   2φ=30⇒φ=15

BDBC=sin2ϕsin(1803ϕ)=sin2ϕsin3ϕADBD=sin90sin(903ϕ)=1cos3ϕBD=ADcos3ϕADcos3ϕAD=sin2ϕsin3ϕ[BC=AD]sin3ϕcos3ϕ=sin2ϕ2sin3ϕcos3ϕ=2sin2ϕsin6ϕ=2sin2ϕ3sin2ϕ4sin32ϕ=2sin2ϕ34sin22ϕ=2[ϕ0]sin2ϕ=122ϕ=30ϕ=15

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