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Question Number 137327 by mr W last updated on 01/Apr/21

$$intriangle\mathrm{\Delta }ABC:BC=1,\mathrm{\angle }B=2\mathrm{\angle }A.$$$$findthemaximumareaof\mathrm{\Delta }ABC.$$

Answered by EDWIN88 last updated on 01/Apr/21

$$\mathrm{\angle }\mathrm{A}+\mathrm{\angle }\mathrm{B}+\mathrm{\angle }\mathrm{C}=\pi ;3\mathrm{\angle }\mathrm{A}+\mathrm{\angle }\mathrm{C}=\pi$$$$\mathrm{let}\mathrm{\angle }\mathrm{A}=\alpha ;\mathrm{\angle }\mathrm{B}=2\alpha ;\mathrm{\angle }\mathrm{C}=\pi -3\alpha$$$$\frac{\mathrm{BC}}{\sin \alpha }=\frac{\mathrm{AC}}{\sin 2\alpha }\Rightarrow \frac{1}{\sin \alpha }=\frac{\mathrm{AC}}{2\sin \alpha \cos \alpha }$$$$\mathrm{AC}=2\cos \alpha$$$$\mathrm{Area}\mathrm{of}\mathrm{\Delta }\mathrm{ABC}=\frac{1}{2}.1.\mathrm{AC}.\sin (\pi -3\alpha )$$$$\mathrm{A}\left(\alpha \right)=\frac{1}{2}\left(2\cos \alpha \sin 3\alpha \right)$$$$\mathrm{A}\left(\alpha \right)=\frac{1}{2}(\sin 4\alpha +\sin 2\alpha )$$$${\mathrm{A}}^{\prime }\left(\alpha \right)=\frac{1}{2}(4\cos 4\alpha +2\cos 2\alpha )=0$$$$\Rightarrow 2\cos 4\alpha +\cos 2\alpha =0$$$$\Rightarrow 2(2\cos {}^{2}2\alpha -1)+\cos 2\alpha =0$$$$\Rightarrow 4\cos {}^{2}2\alpha +\cos 2\alpha -2=0$$$$\cos 2\alpha =\frac{-1+\sqrt{33}}{8}=0.593$$$$\sin 2\alpha =\sqrt{1-{(0.593)}^2}=0.805$$$$\mathrm{Thus}\mathrm{A}{\left(\alpha \right)}_{\max }=\frac{1}{2}(2\sin 2\alpha \cos 2\alpha +\sin 2\alpha )$$$$\mathrm{A}{\left(\alpha \right)}_{\max }=\frac{1}{2}\sin 2\alpha (2\cos 2\alpha +1)$$$$\mathrm{A}{\left(\alpha \right)}_{\max }=\frac{1}{2}(0.805)(2.186)=0.8798$$

Commented by mr W last updated on 01/Apr/21

$$yes.theexactvalueis$$$$\frac{(\sqrt{33}+3)\sqrt{(\sqrt{33}+3)(\sqrt{33}-1)}}{64}\approx 0.8801$$

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