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Question Number 177769 by a.lgnaoui last updated on 08/Oct/22

determiner les elements demandes dans le graphe

$${determiner}\:{les}\:{elements}\:{demandes} \\ $$$${dans}\:{le}\:{graphe} \\ $$

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

$$\measuredangle\mathrm{ADB}=\mathrm{180}−\left(\mathrm{65}+\mathrm{70}\right)=\mathrm{45} \\ $$$$\measuredangle\mathrm{BDC}=\mathrm{45}\:\:\:\:\Rightarrow\measuredangle\mathrm{ADC}=\mathrm{90} \\ $$$$\mathrm{Donc}\:\:\:\bigtriangleup\mathrm{ADC}\:\mathrm{triangle}\:\mathrm{rectangle} \\ $$$$\mathrm{doonc}\:\:\mathrm{AC}\:\:\mathrm{passe}\:\mathrm{par}\:\mathrm{centre}\:\mathrm{du}\:\mathrm{cercle} \\ $$$$\mathrm{AC}=\mathrm{diametre}\: \\ $$$$\mathrm{ACB}\:\:\mathrm{aussi}\:\mathrm{triangle}\:\mathrm{rectangle} \\ $$$$\mathrm{donc}\:\:\measuredangle\mathrm{DBC}=\mathrm{90}−\mathrm{70}=\mathrm{20} \\ $$$$\mathrm{Donc}\:\measuredangle\:\mathrm{BCD}=\mathrm{180}−\left(\mathrm{45}+\mathrm{20}\right)=\mathrm{115} \\ $$

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°

$$\Delta\mathrm{OAB}\:\equiv\:\Delta\mathrm{OCD}\: \\ $$$$\Rightarrow\angle\mathrm{OCD}=\angle\mathrm{OBA}=\mathrm{70}° \\ $$$$\Rightarrow\angle\mathrm{OAB}=\angle\mathrm{ODC}=\mathrm{45}° \\ $$$$\Rightarrow\angle\mathrm{OAD}=\mathrm{65}°−\angle\mathrm{OAB}=\mathrm{20}° \\ $$$$\Rightarrow\angle\mathrm{AOB}=\mathrm{180}°−\mathrm{70}°−\mathrm{45}°=\mathrm{65}° \\ $$$$\Rightarrow\angle\mathrm{BOC}=\mathrm{180}°−\angle\mathrm{AOB}=\mathrm{115}° \\ $$$$\Delta\mathrm{OBC}\:\equiv\:\Delta\mathrm{OAD} \\ $$$$\Rightarrow\angle\mathrm{OBC}=\angle\mathrm{OAD}=\mathrm{20}° \\ $$$$\Rightarrow\angle\mathrm{OCB}=\mathrm{180}°−\angle\mathrm{BOC}−\angle\mathrm{OBC}=\mathrm{45}° \\ $$$$\left({i}\right)\:\angle\mathrm{BCD}=\angle\mathrm{OCB}+\angle\mathrm{OCD}=\mathrm{45}°+\mathrm{70}°=\mathrm{115}° \\ $$$$\left({ii}\right)\:\angle\mathrm{ADB}=\angle\mathrm{OCB}=\mathrm{45}° \\ $$

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