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Category: Geometry

Question-181581

Question Number 181581 by HeferH last updated on 27/Nov/22 Answered by a.lgnaoui last updated on 27/Nov/22 ACDADC=162sin(162)=sin18ACsin18=CDsin12(1)BDC=18DCB=162αBDsin(162α)=CDsinα(BD=AC)$$\frac{\mathrm{AC}}{\mathrm{sin}\:\left(\mathrm{162}−\alpha\right)}=\frac{\mathrm{CD}}{\mathrm{sin}\:\alpha}\:\:\left(\mathrm{2}\right)…