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Question Number 177769 by a.lgnaoui last updated on 08/Oct/22
determiner les elements demandes dans le graphe
determinerleselementsdemandesdanslegraphe
Commented by a.lgnaoui last updated on 08/Oct/22
Commented by a.lgnaoui last updated on 09/Oct/22
∡ADB=180−(65+70)=45 ∡BDC=45 ⇒∡ADC=90 Donc △ADC triangle rectangle doonc AC passe par centre du cercle AC=diametre ACB aussi triangle rectangle donc ∡DBC=90−70=20 Donc ∡ BCD=180−(45+20)=115
∡ADB=180−(65+70)=45∡BDC=45⇒∡ADC=90Donc△ADCtrianglerectangledooncACpasseparcentreducercleAC=diametreACBaussitrianglerectangledonc∡DBC=90−70=20Donc∡BCD=180−(45+20)=115
Answered by Ar Brandon last updated on 08/Oct/22
ΔOAB ≡ ΔOCD ⇒∠OCD=∠OBA=70° ⇒∠OAB=∠ODC=45° ⇒∠OAD=65°−∠OAB=20° ⇒∠AOB=180°−70°−45°=65° ⇒∠BOC=180°−∠AOB=115° ΔOBC ≡ ΔOAD ⇒∠OBC=∠OAD=20° ⇒∠OCB=180°−∠BOC−∠OBC=45° (i) ∠BCD=∠OCB+∠OCD=45°+70°=115° (ii) ∠ADB=∠OCB=45°
ΔOAB≡ΔOCD⇒∠OCD=∠OBA=70°⇒∠OAB=∠ODC=45°⇒∠OAD=65°−∠OAB=20°⇒∠AOB=180°−70°−45°=65°⇒∠BOC=180°−∠AOB=115°ΔOBC≡ΔOAD⇒∠OBC=∠OAD=20°⇒∠OCB=180°−∠BOC−∠OBC=45°(i)∠BCD=∠OCB+∠OCD=45°+70°=115°(ii)∠ADB=∠OCB=45°

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