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Question Number 172408 by infinityaction last updated on 26/Jun/22

Answered by mr W last updated on 26/Jun/22

Commented by mr W last updated on 26/Jun/22

$$AD=DC=1$$$$CP=1\sin \theta =PQ$$$$DQ=1\cos \theta +1\sin \theta =\sin \theta +\cos \theta$$$${AQ}^2=1^2+{(\sin \theta +\cos \theta )}^2-2\times 1\times (\sin \theta +\cos \theta )\cos (\frac{\pi }{2}+\theta )$$$$=2+\sin 2\theta +2(\sin \theta +\cos \theta )\sin \theta$$$$=3+2\sin 2\theta -\cos 2\theta$$$$=3+\sqrt{5}(\frac{2}{\sqrt{5}}\sin 2\theta -\frac{1}{\sqrt{5}}\cos 2\theta )$$$$=3+\sqrt{5}(\cos \alpha \sin 2\theta -\sin \alpha \cos 2\theta )$$$$=3+\sqrt{5}\sin (2\theta -\alpha )$$$$(with\alpha ={\sin }^{-1}\frac{1}{\sqrt{5}}={\cos }^{-1}\frac{2}{\sqrt{5}}={\tan }^{-1}\frac{1}{2})$$$$=3+\sqrt{5}\sin (2\theta -{\tan }^{-1}\frac{1}{2})$$$${\left({AQ}^2\right)}_{max}=3+\sqrt{5}$$$${AQ}_{max}=\sqrt{3+\sqrt{5}}\approx 2.288$$$$at2\theta -{\tan }^{-1}\frac{1}{2}=\frac{\pi }{2}$$$$or\theta =\frac{\pi }{4}+\frac{1}{2}{\tan }^{-1}\frac{1}{2}\approx 58.3°$$

Commented by infinityaction last updated on 26/Jun/22

$$explainsir$$$$3+\sqrt{5}\sin (2\theta -{\tan }^{-1}\frac{1}{2})$$

Commented by mr W last updated on 26/Jun/22

$$ihaveaddedsomestepsmore.$$

Commented by infinityaction last updated on 26/Jun/22

$$AD=DC=1$$$$CP=1\sin \theta =PQ$$$$DQ=1\cos \theta +1\sin \theta =\sin \theta +\cos \theta$$$${\mathit{A}\mathit{Q}}^2=1^2+{(\mathbf{sin}\mathit{\theta }+\mathbf{cos}\mathit{\theta })}^2-2(\mathbf{sin}\mathit{\theta }+\mathbf{cos}\mathit{\theta })\mathbf{cos}(\frac{\mathit{\pi }}{2}+\mathit{\theta })$$$${AQ}^2=2+\sin 2\theta +2(\sin \theta +\cos \theta )\sin \theta$$$${AQ}^2=3+2\sin 2\theta -\cos 2\theta$$$$letp=2\sin 2\theta -\cos 2\theta$$$$\frac{dp}{d\theta }=4\cos 2\theta +2\sin 2\theta =0$$$$\tan 2\theta =-2$$$$\sin 2\theta =\frac{2}{\sqrt{5}}and\cos 2\theta =-\frac{1}{\sqrt{5}}$$$${AQ}^2=3+\sqrt{5}$$$$AQ=\sqrt{3+\sqrt{5}}$$

Commented by infinityaction last updated on 26/Jun/22

$$thankyousir$$

Commented by Tawa11 last updated on 26/Jun/22

$$\mathrm{Great}\mathrm{sir}$$

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