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Question Number 193938 by yaslm last updated on 23/Jun/23

Answered by BaliramKumar last updated on 23/Jun/23

$$15°$$

Commented by yaslm last updated on 23/Jun/23

write method

Answered by mr W last updated on 23/Jun/23

$$\sqrt{2}a=\sqrt{{(a+b)}^2+b^2}$$$$a^2=2ab+2b^2$$$${\left(\frac{a}{b}\right)}^2-2\left(\frac{a}{b}\right)-2=0$$$$\Rightarrow \frac{a}{b}=1+\sqrt{3}$$$$\tan \alpha =\frac{b}{a+b}=\frac{1}{1+\sqrt{3}+1}=2-\sqrt{3}$$$$\tan 2\alpha =\frac{2(2-\sqrt{3})}{1-{(2-\sqrt{3})}^2}=\frac{2-\sqrt{3}}{-3+2\sqrt{3}}=\frac{1}{\sqrt{3}}$$$$\Rightarrow 2\alpha =30°$$$$\Rightarrow \alpha =15°$$

Answered by BaliramKumar last updated on 23/Jun/23

$$\mathrm{Let}\mathrm{AB}=\mathrm{x}\mathrm{\&}\mathrm{BE}=\mathrm{FE}=\mathrm{y}\Rightarrow \mathrm{AC}=\mathrm{AF}=\sqrt{2}\mathrm{x}$$$${\mathrm{AF}}^2={\mathrm{AE}}^2+{\mathrm{FE}}^2$$$${\left(\sqrt{2}\mathrm{x}\right)}^2={(\mathrm{x}+\mathrm{y})}^2+{\mathrm{y}}^2$$$${\mathrm{x}}^2-2\mathrm{yx}-{2\mathrm{y}}^2=0$$$$\mathrm{x}=\frac{2\mathrm{y}\pm \sqrt{{12\mathrm{y}}^2}}{2}=\mathrm{y}\pm \sqrt{3}\mathrm{y}$$$$\mathrm{x}=\mathrm{y}+\sqrt{3}\mathrm{y}[\mathrm{x}\ne \mathrm{y}-\sqrt{3}\mathrm{y},\because \mathrm{x}>\mathrm{y}]$$$$\mathrm{x}=(\sqrt{3}+1)\mathrm{y}$$$$\sin \alpha =\frac{\mathrm{FE}}{\mathrm{AF}}=\frac{\mathrm{y}}{\sqrt{2}\left(\mathrm{x}\right)}=\frac{\cancel{\mathrm{y}}}{\sqrt{2}(\sqrt{3}+1)\cancel{\mathrm{y}}}=\frac{\sqrt{3}-1}{\sqrt{2}(3-1)}$$$$\sin \alpha =\frac{\sqrt{3}-1}{2\sqrt{2}}=\sin 15°$$$$\alpha =15°$$

Answered by talminator2856792 last updated on 23/Jun/23

$$\mathrm{by}\mathrm{pythagorean}\mathrm{theorem}:$$$${AB}^2+{BC}^2={AB}^2+{BE}^2+2AB\cdot BE+{FE}^2$$$$$$$${AB}^2={BE}^2+2AB\cdot BE+{FE}^2$$$${AB}^2=2{BE}^2+2AB\cdot BE$$$${AB}^2-2AB\cdot BE-2{BE}^2=0$$$${AB}^2-2AB\cdot BE+{BE}^2=3{BE}^2$$$${(AB-BE)}^2=3{BE}^2$$$${(AB-BE)}^2-{\left(\sqrt{3}BE\right)}^2=0$$$$(AB-BE-\sqrt{3}BE)(AB-BE+\sqrt{3}BE)=0$$$$(AB-BE(1+\sqrt{3}))(AB-BE(1-\sqrt{3}))=0$$$$AB=BE(1+\sqrt{3})$$$$AB=BE(\sqrt{3}-1)$$$$$$$$\Rightarrow \alpha =\arctan \frac{1}{2+\sqrt{3}}$$$$\alpha =15°$$

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