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Question Number 214350 by ajfour last updated on 06/Dec/24

Answered by mr W last updated on 06/Dec/24

$$\tan \alpha =\frac{1}{2}\Rightarrow \tan \frac{\alpha }{2}=\sqrt{5}-2$$$$2+r\tan \frac{\alpha }{2}=\sqrt{{(r+1)}^2-{(r-1)}^2}$$$$2+(\sqrt{5}-2)r=2\sqrt{r}$$$$(\sqrt{5}-2)r-2\sqrt{r}+2=0$$$$\sqrt{r}=\frac{1-\sqrt{5-2\sqrt{5}}}{\sqrt{5}-2}$$$$\Rightarrow r={\left(\frac{1-\sqrt{5-2\sqrt{5}}}{\sqrt{5}-2}\right)}^2\approx 1.342$$

Commented by ajfour last updated on 06/Dec/24

https://youtu.be/6veUu8wBV_8?si=aJGNsb2DtsOBFYbJ

Commented by ajfour last updated on 06/Dec/24

sir could you check the question in the video for me?

Commented by mr W last updated on 06/Dec/24

$$forsomereasoni{can}^{\prime }topenthis$$$$video.$$

Commented by mr W last updated on 06/Dec/24

$$icanviewitnow.great!$$

Answered by A5T last updated on 06/Dec/24

Commented by A5T last updated on 06/Dec/24

$$AF=\sqrt{{(1+r)}^2-r^2}=\sqrt{1+2r}$$$$\Rightarrow FE=\sqrt{1^2+2^2}-\sqrt{1+2r}=\sqrt{5}-\sqrt{1+2r}=EC$$$${(1+r)}^2={(r-1)}^2+{(2+\sqrt{5}-\sqrt{1+2r})}^2$$$$\Rightarrow r=14+6\sqrt{5}-2\sqrt{85+38\sqrt{5}}\approx 1.3419$$

Commented by ajfour last updated on 06/Dec/24

$$Thanks.Isee.$$

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