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Question Number 186079 by TUN last updated on 31/Jan/23
Answered by a.lgnaoui last updated on 01/Feb/23
△ABDet△ACDAD:cotecommun△ABD:sin46BD=sinXAD(1)△ACDsin12CD=sin62AD(2)(1)et(2)⇒BDsinXsin46=CDsin62sin12(3)△BCD∡CDB=52−Xsin8BD=sin(52−X)CD(4)(3)×(4)⇒sin8×sinXsin46=sin62×sin(52−X)sin12sin8×sin12×sinX=sin62×sin46sin(52−X)sin(52−X)=sin521−sin2X−cos52sinXposinsZ=sinXsinXsin(52−X)=sin62×sin46sin8×sin12=sinXsin52cosX−cos52sinX=sin62×sin46sin8×sin12=(tanX)(sin52−cos52tanX)=sin62×sin46sin8×sin12posinst=tanXtsin52−tcos52=sin62×sin46sin8×sin12t(sin8sin12+sin62sin46cos52)=sin52sin62sin46)t=sin52sin62sin46sin8sin12+sin62sin46cos52t=1,191753592594X=50°
Commented by a.lgnaoui last updated on 01/Feb/23
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