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Question Number 182407 by Acem last updated on 09/Dec/22

Answered by a.lgnaoui last updated on 09/Dec/22

$$Area=\mathrm{AC}\times \mathrm{AB}\frac{\sin \beta }{2}$$$$\beta =\frac{12}{5}\mathrm{x}\gamma =\frac{3\mathrm{x}}{2}\alpha =\frac{3}{5}\mathrm{x}\left(degre\right)$$$$\alpha +\beta +\gamma =180\left(\frac{45}{10}\right)\times \mathrm{x}=180\mathrm{x}=40$$$$\beta =\frac{12\times 40}{5}=96°$$$$Area=\frac{2\times \mathrm{AB}}{2}\times \sin 96$$$$△\mathrm{ABC}\frac{\sin \alpha }{\mathrm{AB}}=\frac{\sin \gamma }{2}=\frac{\sin 60}{2}=\frac{\sqrt{3}}{4}$$$$4\sin \alpha =\mathrm{AB}\times \sqrt{3}\mathrm{AB}=\frac{4\sqrt{3}\times \sin \left(24\right)}{3}$$$$\alpha =\frac{3}{5}\times 40=24°\mathrm{AB}=0,939318$$$$Area=\mathrm{AB}\times \sin 96$$$$$$$$Area=0,9342$$$$$$

Commented by Acem last updated on 10/Dec/22

$$Tousmercietrespect!$$

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