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Question Number 173026 by mnjuly1970 last updated on 05/Jul/22

Answered by mr W last updated on 05/Jul/22

Commented by mr W last updated on 05/Jul/22

$$b=r_2+\frac{r_2}{\tan \frac{\pi -\beta -\gamma }{2}}=r_2(1+\tan \frac{\beta +\gamma }{2})$$$$r_2=\frac{b}{1+\tan \frac{\beta +\gamma }{2}}$$$$similarly$$$$r_1=\frac{a}{1+\tan \frac{\alpha +\delta }{2}}$$$$\beta +\gamma =\pi -(\alpha +\delta )$$$$\tan \frac{\beta +\gamma }{2}=\tan (\frac{\pi }{2}-\frac{\alpha +\delta }{2})=\frac{1}{\tan \frac{\alpha +\delta }{2}}$$$$r_1+r_2=\frac{b}{1+\frac{1}{\tan \frac{\alpha +\delta }{2}}}+\frac{a}{1+\tan \frac{\alpha +\delta }{2}}$$$$r_1+r_2=\frac{a+b\tan \frac{\alpha +\delta }{2}}{1+\tan \frac{\alpha +\delta }{2}}$$$$r_1+r_2=\frac{g(\cos \alpha +\sin \alpha \tan \frac{\alpha +\delta }{2})}{1+\tan \frac{\alpha +\delta }{2}}$$$$r_1+r_2=\frac{g\cos \frac{\alpha -\delta }{2}}{(1+\tan \frac{\alpha +\delta }{2})\cos \frac{\alpha +\delta }{2}}$$$$similarly$$$$r_3+r_4=\frac{c+d\tan \frac{\beta +\gamma }{2}}{1+\tan \frac{\beta +\gamma }{2}}$$$$r_3+r_4=\frac{d+c\tan \frac{\alpha +\delta }{2}}{1+\tan \frac{\alpha +\delta }{2}}$$$$r_3+r_4=\frac{g(\cos \delta +\sin \delta \tan \frac{\alpha +\delta }{2})}{1+\tan \frac{\alpha +\delta }{2}}$$$$r_3+r_4=\frac{g\cos \frac{\delta -\alpha }{2}}{(1+\tan \frac{\alpha +\delta }{2})\cos \frac{\alpha +\delta }{2}}$$$$r_3+r_4=\frac{g\cos \frac{\alpha -\delta }{2}}{(1+\tan \frac{\alpha +\delta }{2})\cos \frac{\alpha +\delta }{2}}=r_1+r_2✓$$

Commented by Tawa11 last updated on 06/Jul/22

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

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