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Question Number 88886 by I want to learn more last updated on 13/Apr/20

Commented by I want to learn more last updated on 13/Apr/20

$$\mathrm{ABCD}\mathrm{is}\mathrm{a}\mathrm{square},\mathrm{find}\mathrm{AB}.$$$$\mathrm{Please}\mathrm{help}$$

Commented by I want to learn more last updated on 13/Apr/20

$$\mathrm{Thanks}\mathrm{sir}$$

Commented by john santu last updated on 14/Apr/20

$$AFisthebisectoroftheangle$$$$DAE$$

Commented by mr W last updated on 14/Apr/20

$$x=sidelengthofsquare$$$$\alpha +2\beta =\frac{\pi }{2}\Rightarrow \beta =\frac{\pi }{4}-\frac{\alpha }{2}$$$$\tan \alpha =\frac{36}{x}$$$$\tan \beta =\frac{64}{x}=\frac{16}{9}\tan \alpha$$$$\frac{1-\tan \frac{\alpha }{2}}{1+\tan \frac{\alpha }{2}}=\frac{16}{9}\times \frac{2\tan \frac{\alpha }{2}}{1-{\tan }^2\frac{\alpha }{2}}$$$$1-\tan \frac{\alpha }{2}=\frac{16}{9}\times \frac{2\tan \frac{\alpha }{2}}{1-\tan \frac{\alpha }{2}}$$$$1-2\tan \frac{\alpha }{2}+{\tan }^2\frac{\alpha }{2}=\frac{32}{9}\tan \frac{\alpha }{2}$$$$9-50\tan \frac{\alpha }{2}+9{\tan }^2\frac{\alpha }{2}=0$$$$\tan \frac{\alpha }{2}=\frac{25-4\sqrt{34}}{9}\Rightarrow \tan \alpha =\frac{2\times \frac{25-4\sqrt{34}}{9}}{1-{\left(\frac{25-4\sqrt{34}}{9}\right)}^2}=\frac{9\sqrt{34}}{136}$$$$x=\frac{36\times 136}{9\sqrt{34}}=16\sqrt{34}\approx 93.295$$

Commented by I want to learn more last updated on 17/Apr/20

$$\mathrm{Thanks}\mathrm{sir}$$

Answered by john santu last updated on 14/Apr/20

$$AE=100$$$$thenAB=\sqrt{{100}^2-{36}^2}$$$$=\sqrt{136\times 64}=8\sqrt{136}$$$$=16\sqrt{34}\leftarrow theanswer$$

Commented by I want to learn more last updated on 17/Apr/20

$$\mathrm{How}\mathrm{is}\mathrm{AE}100\mathrm{sir}$$

Commented by mr W last updated on 18/Apr/20

Commented by mr W last updated on 18/Apr/20

$$\mathrm{\Delta }EAF{}^{\prime }isisosceles.$$$$AE=EF{}^{\prime }=36+64=100$$

Commented by I want to learn more last updated on 19/Apr/20

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}.$$

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