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Question Number 172144 by mnjuly1970 last updated on 23/Jun/22

Answered by infinityaction last updated on 23/Jun/22

Commented by infinityaction last updated on 23/Jun/22

$$\cos \alpha =\frac{2}{r}and\sin \alpha =\frac{r}{6}$$$${\cos }^2\alpha +{\sin }^2\alpha =\frac{4}{r^2}+\frac{r^2}{36}$$$$let{r}^2=p$$$$1=\frac{4}{p}+\frac{p}{36}$$$$p^2-36p+144=0$$$$p=r^2=18-6\sqrt{5}$$

Commented by mr W last updated on 23/Jun/22

$$nicesolution!$$

Commented by mnjuly1970 last updated on 23/Jun/22

$$thankyousomuch...$$

Answered by mr W last updated on 23/Jun/22

Commented by mr W last updated on 23/Jun/22

$$yes,ifoundittoo.thanks!$$

Commented by mr W last updated on 23/Jun/22

$$\gamma =90°-\alpha =45°+\beta$$$$\frac{\sin \alpha }{d}=\frac{\sin 45°}{6}...\left(i\right)$$$$\frac{d}{\sin 45°}=\frac{4}{\sin (45°+\beta )}=\frac{4}{\cos \alpha }...\left(ii\right)$$$$\left(i\right)\times \left(ii\right):$$$$\frac{\sin \alpha }{\sin 45°}=\frac{4\sin 45°}{6\cos \alpha }$$$$\Rightarrow 6\sin \alpha \cos \alpha =4{\sin }^{2}45°$$$$\Rightarrow 3\sin 2\alpha =2\Rightarrow \sin 2\alpha =\frac{2}{3}$$$$\cos 2\alpha =2{\cos }^2\alpha -1=\sqrt{1-{\left(\frac{2}{3}\right)}^2}=\frac{\sqrt{5}}{3}$$$$\Rightarrow \cos \alpha =\frac{\sqrt{3+\sqrt{5}}}{\sqrt{6}}$$$$d=\frac{4\sin 45°}{\cos \alpha }=\frac{4\sqrt{3}}{\sqrt{3+\sqrt{5}}}$$$$areaofsquare$$$$={\left(\frac{d}{\sqrt{2}}\right)}^2=\frac{d^2}{2}=\frac{8\times 3}{3+\sqrt{5}}=6(3-\sqrt{5})$$

Commented by mnjuly1970 last updated on 23/Jun/22

$$thanksalotsir$$

Commented by infinityaction last updated on 23/Jun/22

$$\frac{d^2}{2}=\frac{1}{2}\times \frac{16\times 3}{3+\sqrt{5}}=6(3-\sqrt{5})=18-6\sqrt{5}$$

Commented by Tawa11 last updated on 24/Jun/22

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

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