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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
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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