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Category: Geometry

Question-184398

Question Number 184398 by HeferH last updated on 06/Jan/23 Answered by som(math1967) last updated on 06/Jan/23 BDBC=sin2ϕsin(1803ϕ)=sin2ϕsin3ϕADBD=sin90sin(903ϕ)=1cos3ϕBD=ADcos3ϕADcos3ϕAD=sin2ϕsin3ϕ[BC=AD]$${sin}\mathrm{3}\phi{cos}\mathrm{3}\phi={sin}\mathrm{2}\phi…