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

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

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

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