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Question Number 186298 by TUN last updated on 03/Feb/23

Answered by a.lgnaoui last updated on 05/Feb/23

$$△ABC:6+36+x+y+\mathrm{z}+12=180$$$$x+y+\mathrm{z}=126$$$$△\mathrm{ACD}:\frac{\sin x}{\mathrm{AD}}=\frac{\sin \mathrm{y}}{\mathrm{CD}}=\frac{\sin (\mathrm{x}+\mathrm{y})}{\mathrm{AC}}\left(1\right)$$$$△\mathrm{ABC}\frac{\sin (\mathrm{y}+12)}{\mathrm{BC}}=\frac{\sin 42}{\mathrm{AC}}\left(2\right)$$$$\left(1\right)\Rightarrow \sin y=\frac{CD}{AD}\sin \mathrm{x}$$$$\mathrm{ACsin}\mathrm{y}=\mathrm{CDsin}(\mathrm{x}+\mathrm{y})\left(3\right)$$$$△\mathrm{BDC}\frac{\sin \mathrm{z}}{\mathrm{BD}}=\frac{\sin (36+\mathrm{z})}{\mathrm{BC}}$$$$\mathrm{BC}=\mathrm{BD}\Rightarrow \sin \mathrm{z}=\sin (36+\mathrm{z})$$$$\pi -\mathrm{z}=36+\mathrm{z}\mathrm{z}=72$$$$(z=\measuredangle DCB=\measuredangle BDC):BD=AC$$$$$$$$\Rightarrow x+y=54$$$$$$$$\frac{\sin 12}{\mathrm{BD}}=\frac{\sin 6}{\mathrm{AD}}\frac{\mathrm{AD}}{\mathrm{BD}}\Rightarrow \frac{\sin 6}{\sin 12}=\frac{\mathrm{AD}}{\mathrm{AC}}\left(4\right)$$$$$$$$\left(1\right)\frac{\sin x}{\mathrm{AD}}=\frac{\sin (\mathrm{x}+\mathrm{y})}{\mathrm{AC}}=\frac{\sin 54}{\mathrm{AC}}$$$$\frac{\mathrm{AC}}{\mathrm{AD}}=\frac{\sin 54}{\sin \mathrm{x}}\left(5\right)$$$$\left(4\right)er\left(5\right)\Rightarrow \frac{\sin x}{\sin 54}=\frac{\sin 6}{\sin 12}$$$$$$$$\sin x=\frac{\sin 54\sin 6}{\sin 12}$$$$$$$$\sin x=0,4067366$$$$\mathrm{X}=24°$$$$$$

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