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Question Number 193027 by Mingma last updated on 02/Jun/23

Answered by BaliramKumar last updated on 02/Jun/23

$$\alpha ={\tan }^{-1}\left(2\sqrt{2}\right)?$$

Answered by ajfour last updated on 02/Jun/23

$$(1+r)\cos \alpha +\frac{r}{\sqrt{2}}=1$$$$(1+r)\sin \alpha +\frac{r}{\sqrt{2}}=2$$$$\tan \alpha =\frac{2-\frac{r}{\sqrt{2}}}{1-\frac{r}{\sqrt{2}}}$$$${(1+r)}^2={(1-\frac{r}{\sqrt{2}})}^2+{(2-\frac{r}{\sqrt{2}})}^2$$$$\Rightarrow 2r+1=5-\frac{6r}{\sqrt{2}}$$$$r=\frac{2\sqrt{2}}{3+\sqrt{2}}$$$$\tan \alpha =\frac{2-\frac{2}{3+\sqrt{2}}}{1-\frac{2}{3+\sqrt{2}}}=\frac{4+2\sqrt{2}}{1+\sqrt{2}}=2\sqrt{2}$$$$\alpha ={\tan }^{-1}\left(2\sqrt{2}\right)$$

Commented by Mingma last updated on 03/Jun/23

Perfect ��

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