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Question Number 148540 by Jonathanwaweh last updated on 29/Jul/21

Answered by Olaf_Thorendsen last updated on 29/Jul/21

$$\mathrm{Formule}{\mathrm{d}}^{\prime }\mathrm{Al}-\mathrm{Kashi}(\mathrm{generalisation}$$$$\mathrm{de}\mathrm{Pythagore}\mathrm{dans}\mathrm{un}\mathrm{triangle}$$$$\mathrm{quelconque}):$$$$c^2=a^2+b^2-2ab\cos \alpha$$$$\Rightarrow {(a-b\cos \alpha )}^2-b^2{\cos }^2\alpha +b^2=c^2$$$${(a-b\cos \alpha )}^2+b^2{\sin }^2\alpha =c^2$$$$a-b\cos \alpha =\pm \sqrt{c^2-b^2{\sin }^2\alpha }$$$$\mathrm{On}\mathrm{verifie}\mathrm{que}\mathrm{le}\mathrm{bon}\mathrm{signe}\mathrm{est}+\mathrm{avec}$$$$\mathrm{une}\mathrm{valeur}\mathrm{particuliere},\alpha =\frac{\pi }{2}\mathrm{par}$$$$\mathrm{exemple}.$$$$a-b\cos \alpha =\sqrt{c^2-b^2{\sin }^2\alpha }$$$$$$$$a=b\cos \alpha +\sqrt{c^2-b^2{\sin }^2\alpha }$$$$\mathrm{La}\mathrm{bonne}\mathrm{reponse}\mathrm{est}\left(\mathrm{a}\right).$$

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