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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(180−3ϕ)=sin2ϕsin3ϕADBD=sin90sin(90−3ϕ)=1cos3ϕBD=ADcos3ϕADcos3ϕAD=sin2ϕsin3ϕ[∵BC=AD]sin3ϕcos3ϕ=sin2ϕ2sin3ϕcos3ϕ=2sin2ϕsin6ϕ=2sin2ϕ3sin2ϕ−4sin32ϕ=2sin2ϕ⇒3−4sin22ϕ=2[ϕ≠0]sin2ϕ=122ϕ=30⇒ϕ=15

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