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Question Number 168906 by cortano1 last updated on 21/Apr/22

Answered by mr W last updated on 21/Apr/22

$$\tan \frac{B}{2}=\frac{\frac{b}{c}}{1+\frac{a}{c}}=\frac{b}{a+c}=\frac{b}{a+\sqrt{a^2+b^2}}$$$$\tan \frac{A}{2}=\frac{\frac{a}{c}}{1+\frac{b}{c}}=\frac{a}{b+c}=\frac{a}{b+\sqrt{a^2+b^2}}$$$$\frac{r}{\tan \frac{B}{2}}+\frac{r}{\tan \frac{A}{2}}+(n-1)r=c=\sqrt{a^2+b^2}$$$$r[\frac{a+\sqrt{a^2+b^2}}{b}+\frac{b+\sqrt{a^2+b^2}}{a}+n-1]=\sqrt{a^2+b^2}$$$$\Rightarrow r=\frac{\sqrt{a^2+b^2}}{\frac{a+\sqrt{a^2+b^2}}{b}+\frac{b+\sqrt{a^2+b^2}}{a}+n-1}$$

Commented by Tawa11 last updated on 21/Apr/22

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

Commented by cortano1 last updated on 22/Apr/22

$$waw...great$$

Commented by peter frank last updated on 26/Apr/22

$$\mathrm{first},\mathrm{second}\mathrm{line}.\mathrm{why}1+\frac{\mathrm{a}}{\mathrm{c}},1+\frac{\mathrm{b}}{\mathrm{c}}?$$

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

$$\tan \frac{\theta }{2}=\frac{\sin \theta }{1+\cos \theta }$$

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