Question-181581

Question Number 181581 by HeferH last updated on 27/Nov/22

Answered by a.lgnaoui last updated on 27/Nov/22

△ACD ∡ADC=162 sin (162)=sin 18 ((AC)/(sin 18))=((CD)/(sin 12)) (1) ∡BDC=18 ∡DCB=162−α ((BD)/(sin (162−α)))=((CD)/(sin α)) (BD=AC)⇒ ((AC)/(sin (162−α)))=((CD)/(sin α)) (2) (1) et (2) ⇒((sin 18)/(sin 12))=((sin (162−α))/(sin α)) sin 18×sin α=sin 12×sin (162−α) sin α(sin 18−sin 12cos 18)=(sin 12×sin 18)cos α tan α=((sin 12×sin 18)/(sin 18−sin 12×cos 18)) tan α=0,5773502621 α=30°

$$\bigtriangleup\mathrm{ACD}\:\:\:\measuredangle\mathrm{ADC}=\mathrm{162}\:\:\:\mathrm{sin}\:\left(\mathrm{162}\right)=\mathrm{sin}\:\mathrm{18} \\ $$$$\frac{\mathrm{AC}}{\mathrm{sin}\:\mathrm{18}}=\frac{\mathrm{CD}}{\mathrm{sin}\:\mathrm{12}}\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\measuredangle\mathrm{BDC}=\mathrm{18}\:\:\measuredangle\mathrm{DCB}=\mathrm{162}−\alpha \\ $$$$\frac{\mathrm{BD}}{\mathrm{sin}\:\left(\mathrm{162}−\alpha\right)}=\frac{\mathrm{CD}}{\mathrm{sin}\:\alpha}\:\left(\mathrm{BD}=\mathrm{AC}\right)\Rightarrow \\ $$$$\frac{\mathrm{AC}}{\mathrm{sin}\:\left(\mathrm{162}−\alpha\right)}=\frac{\mathrm{CD}}{\mathrm{sin}\:\alpha}\:\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)\:{et}\:\left(\mathrm{2}\right)\:\:\Rightarrow\frac{\mathrm{sin}\:\mathrm{18}}{\mathrm{sin}\:\mathrm{12}}=\frac{\mathrm{sin}\:\left(\mathrm{162}−\alpha\right)}{\mathrm{sin}\:\alpha} \\ $$$$\mathrm{sin}\:\mathrm{18}×\mathrm{sin}\:\alpha=\mathrm{sin}\:\mathrm{12}×\mathrm{sin}\:\left(\mathrm{162}−\alpha\right) \\ $$$$\mathrm{sin}\:\alpha\left(\mathrm{sin}\:\mathrm{18}−\mathrm{sin}\:\mathrm{12cos}\:\mathrm{18}\right)=\left(\mathrm{sin}\:\mathrm{12}×\mathrm{sin}\:\mathrm{18}\right)\mathrm{cos}\:\alpha \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{sin}\:\mathrm{12}×\mathrm{sin}\:\mathrm{18}}{\mathrm{sin}\:\mathrm{18}−\mathrm{sin}\:\mathrm{12}×\mathrm{cos}\:\mathrm{18}} \\ $$$$\mathrm{tan}\:\alpha=\mathrm{0},\mathrm{5773502621} \\ $$$$\:\:\alpha=\mathrm{30}° \\ $$$$ \\ $$

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