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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 +(r_2 /(tan ((π−β−γ)/2)))=r_2 (1+tan ((β+γ)/2)) r_2 =(b/(1+tan ((β+γ)/2))) similarly r_1 =(a/(1+tan ((α+δ)/2))) β+γ=π−(α+δ) tan ((β+γ)/2)=tan ((π/2)−((α+δ)/2))=(1/(tan ((α+δ)/2))) r_1 +r_2 =(b/(1+(1/(tan ((α+δ)/2)))))+(a/(1+tan ((α+δ)/2))) r_1 +r_2 =((a+b tan ((α+δ)/2))/(1+tan ((α+δ)/2))) r_1 +r_2 =((g(cos α+sin α tan ((α+δ)/2)))/(1+tan ((α+δ)/2))) r_1 +r_2 =((g cos ((α−δ)/2))/((1+tan ((α+δ)/2))cos ((α+δ)/2))) similarly r_3 +r_4 =((c+d tan ((β+γ)/2))/(1+tan ((β+γ)/2))) r_3 +r_4 =((d+c tan ((α+δ)/2))/(1+tan ((α+δ)/2))) r_3 +r_4 =((g(cos δ+sin δ tan ((α+δ)/2)))/(1+tan ((α+δ)/2))) r_3 +r_4 =((g cos ((δ−α)/2))/((1+tan ((α+δ)/2))cos ((α+δ)/2))) r_3 +r_4 =((g cos ((α−δ)/2))/((1+tan ((α+δ)/2))cos ((α+δ)/2)))=r_1 +r_2 ✓

$${b}={r}_{\mathrm{2}} +\frac{{r}_{\mathrm{2}} }{\mathrm{tan}\:\frac{\pi−\beta−\gamma}{\mathrm{2}}}={r}_{\mathrm{2}} \left(\mathrm{1}+\mathrm{tan}\:\frac{\beta+\gamma}{\mathrm{2}}\right) \\ $$$${r}_{\mathrm{2}} =\frac{{b}}{\mathrm{1}+\mathrm{tan}\:\frac{\beta+\gamma}{\mathrm{2}}} \\ $$$${similarly} \\ $$$${r}_{\mathrm{1}} =\frac{{a}}{\mathrm{1}+\mathrm{tan}\:\frac{\alpha+\delta}{\mathrm{2}}} \\ $$$$\beta+\gamma=\pi−\left(\alpha+\delta\right) \\ $$$$\mathrm{tan}\:\frac{\beta+\gamma}{\mathrm{2}}=\mathrm{tan}\:\left(\frac{\pi}{\mathrm{2}}−\frac{\alpha+\delta}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\alpha+\delta}{\mathrm{2}}} \\ $$$${r}_{\mathrm{1}} +{r}_{\mathrm{2}} =\frac{{b}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\alpha+\delta}{\mathrm{2}}}}+\frac{{a}}{\mathrm{1}+\mathrm{tan}\:\frac{\alpha+\delta}{\mathrm{2}}} \\ $$$${r}_{\mathrm{1}} +{r}_{\mathrm{2}} =\frac{{a}+{b}\:\mathrm{tan}\:\frac{\alpha+\delta}{\mathrm{2}}}{\mathrm{1}+\mathrm{tan}\:\frac{\alpha+\delta}{\mathrm{2}}} \\ $$$${r}_{\mathrm{1}} +{r}_{\mathrm{2}} =\frac{{g}\left(\mathrm{cos}\:\alpha+\mathrm{sin}\:\alpha\:\mathrm{tan}\:\frac{\alpha+\delta}{\mathrm{2}}\right)}{\mathrm{1}+\mathrm{tan}\:\frac{\alpha+\delta}{\mathrm{2}}} \\ $$$${r}_{\mathrm{1}} +{r}_{\mathrm{2}} =\frac{{g}\:\mathrm{cos}\:\frac{\alpha−\delta}{\mathrm{2}}}{\left(\mathrm{1}+\mathrm{tan}\:\frac{\alpha+\delta}{\mathrm{2}}\right)\mathrm{cos}\:\frac{\alpha+\delta}{\mathrm{2}}} \\ $$$${similarly} \\ $$$${r}_{\mathrm{3}} +{r}_{\mathrm{4}} =\frac{{c}+{d}\:\mathrm{tan}\:\frac{\beta+\gamma}{\mathrm{2}}}{\mathrm{1}+\mathrm{tan}\:\frac{\beta+\gamma}{\mathrm{2}}} \\ $$$${r}_{\mathrm{3}} +{r}_{\mathrm{4}} =\frac{{d}+{c}\:\mathrm{tan}\:\frac{\alpha+\delta}{\mathrm{2}}}{\mathrm{1}+\mathrm{tan}\:\frac{\alpha+\delta}{\mathrm{2}}} \\ $$$${r}_{\mathrm{3}} +{r}_{\mathrm{4}} =\frac{{g}\left(\mathrm{cos}\:\delta+\mathrm{sin}\:\delta\:\mathrm{tan}\:\frac{\alpha+\delta}{\mathrm{2}}\right)}{\mathrm{1}+\mathrm{tan}\:\frac{\alpha+\delta}{\mathrm{2}}} \\ $$$${r}_{\mathrm{3}} +{r}_{\mathrm{4}} =\frac{{g}\:\mathrm{cos}\:\:\frac{\delta−\alpha}{\mathrm{2}}}{\left(\mathrm{1}+\mathrm{tan}\:\frac{\alpha+\delta}{\mathrm{2}}\right)\mathrm{cos}\:\frac{\alpha+\delta}{\mathrm{2}}} \\ $$$${r}_{\mathrm{3}} +{r}_{\mathrm{4}} =\frac{{g}\:\mathrm{cos}\:\:\frac{\alpha−\delta}{\mathrm{2}}}{\left(\mathrm{1}+\mathrm{tan}\:\frac{\alpha+\delta}{\mathrm{2}}\right)\mathrm{cos}\:\frac{\alpha+\delta}{\mathrm{2}}}={r}_{\mathrm{1}} +{r}_{\mathrm{2}} \:\checkmark \\ $$

Commented by Tawa11 last updated on 06/Jul/22

Great sir

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

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