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Question Number 184557 by a.lgnaoui last updated on 08/Jan/23

$$Determiner$$$$1\bullet \mathrm{AB},\mathrm{BC}\mathrm{AC}\mathrm{en}\mathrm{fonction}\mathrm{de}\mathbf{r}$$$$2\bullet \measuredangle \mathrm{CBA};\measuredangle \mathrm{BAC};et\measuredangle \mathrm{BCA}$$

Commented by a.lgnaoui last updated on 08/Jan/23

Answered by a.lgnaoui last updated on 09/Jan/23

$$$$$$Reponse$$$$-----------------$$$$\measuredangle AOB=60=2\measuredangle ACCB\Rightarrow \measuredangle ACB=30$$$$\left(Proprietedeangleaucentre\right)$$$$1\bullet {AB}^2=2r^2(1-\cos 60)\Rightarrow \mathit{A}\mathit{B}=\mathit{r}(△AOBEquilaterale)$$$${AB}^2={AC}^2+{BC}^2-2AC\times BC\cos 30$$$$AB=\frac{5}{2}BC\Rightarrow \mathit{B}\mathit{C}=\frac{2}{5}\mathit{r}$$$$r^2={AC}^2+\frac{4}{25}{r}^2-\frac{4}{5}rAC\cos 30$$$$r^2={AC}^2+\frac{4}{25}{r}^2-\frac{2}{5}r\sqrt{3}\times AC$$$$AC=x$$$$x^2-\frac{2r\sqrt{3}}{5}x-\frac{21}{25}{r}^2=0$$$$(x-\frac{\sqrt{3}+2\sqrt{6}}{5}r)(x-\frac{\sqrt{3}-2\sqrt{6}}{5}r)=0$$$$AC=\frac{\sqrt{3}+2\sqrt{6}}{5}\mathit{r}$$$$3\bullet caluldesangles$$$$\measuredangle BCA=\frac{60}{2}=30$$$$\frac{\sin \left(\measuredangle ABC\right)}{\left(\frac{\sqrt{3}+2\sqrt{6}}{5}\right)r}=\frac{\sin 30}{r}=\frac{1}{2r}$$$$\sin \left(\measuredangle ABC\right)=\frac{\sqrt{3}+2\sqrt{6}}{10}=0,6631$$$$\measuredangle ABC={\sin }^{-1}\left(\frac{\sqrt{3}+2\sqrt{6}}{10}\right)$$$$\measuredangle \mathit{A}\mathit{B}\mathit{C}=41,53$$$$$$$$\measuredangle BAC=(180-30-41,53)$$$$\measuredangle \mathit{B}\mathit{A}\mathit{C}=108,47$$$$$$

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