Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 174799 by Shrinava last updated on 11/Aug/22

Answered by behi834171 last updated on 12/Aug/22

$$cot\frac{A}{2}=\sqrt{\frac{p(p-a)}{(p-b)(p-c)}}=\frac{\mathit{S}}{(\mathit{p}-\mathit{b})(\mathit{p}-\mathit{c})}=$$$$=\frac{\mathit{p}(\mathit{p}-\mathit{a})}{\mathit{S}}=\frac{\mathit{p}-\mathit{a}}{\mathit{r}}andsoon...$$$$\mathrm{\Pi }\mathit{c}\mathit{o}\mathit{t}\frac{\mathit{A}}{2}=\frac{{\mathit{S}}^2}{\mathit{p}.{\mathit{r}}^3}=\frac{\mathit{S}}{{\mathit{r}}^2}$$$${\left(\mathit{l}\mathit{h}\mathit{s}\right)}^2=\frac{{(\mathit{p}-\mathit{a})}^2}{{\mathit{r}}^2}+\frac{{(\mathit{p}-\mathit{b})}^2}{{\mathit{r}}^2}+4\times \frac{\mathit{p}-\mathit{c}}{\mathit{r}}+$$$$+2\times \frac{(\mathit{p}-\mathit{a})(\mathit{p}-\mathit{b})}{{\mathit{r}}^2}+4\times (\frac{(\mathit{p}-\mathit{a})}{\mathit{r}}+\frac{(\mathit{p}-\mathit{b})}{\mathit{r}}).\sqrt{\frac{\mathit{p}-\mathit{c}}{\mathit{r}}}=$$$$=\left[\frac{{(\mathit{p}-\mathit{a})}^2+{(\mathit{p}-\mathit{b})}^2+2(\mathit{p}-\mathit{a})(\mathit{p}-\mathit{b})}{{\mathit{r}}^2}\right]+$$$$+\frac{4(\mathit{p}-\mathit{c})}{\mathit{r}}+4\times \frac{2\mathit{p}-\mathit{a}-\mathit{b}}{\mathit{r}}\sqrt{\frac{\mathit{p}-\mathit{c}}{\mathit{r}}}=$$$$=\frac{{\mathit{c}}^2}{{\mathit{r}}^2}+4\times \frac{\mathit{c}}{\mathit{r}}\sqrt{\frac{\mathit{p}-\mathit{c}}{\mathit{r}}}+4\times \frac{\mathit{p}-\mathit{c}}{\mathit{r}}=$$$$={(\frac{\mathit{c}}{\mathit{r}}+2\sqrt{\frac{\mathit{p}-\mathit{c}}{\mathit{r}}})}^2=\frac{{(\mathit{c}+2\sqrt{\mathit{S}-\mathit{r}.\mathit{c}})}^2}{{\mathit{r}}^2}$$$${\left(\mathit{l}\mathit{h}\mathit{s}\right)}^4=\frac{{\mathit{c}}^2+4(\mathit{S}-\mathit{r}.\mathit{c})+4\sqrt{\mathit{S}-\mathit{r}.\mathit{c}}}{{\mathit{r}}^4}$$$${\left(\mathit{R}\mathit{H}\mathit{S}\right)}^4=4^4\times \frac{\mathit{S}}{{\mathit{r}}^2}\Rightarrow$$$$\Rightarrow {\mathit{c}}^2+4(\mathit{S}-\mathit{r}.\mathit{c})+4\sqrt{\mathit{S}-\mathit{r}.\mathit{c}}=256\mathit{S}.{\mathit{r}}^2\Rightarrow$$$$\Rightarrow 16(\mathit{S}-\mathit{r}.\mathit{c})={256}^2{\mathit{S}}^2.{\mathit{r}}^4+{\mathit{c}}^4+16{(\mathit{S}-\mathit{r}.\mathit{c})}^2-$$$$-512{\mathit{S}\mathit{r}}^2{\mathit{c}}^2-512\times 8\mathit{S}.{\mathit{r}}^2.(\mathit{S}-\mathit{r}.\mathit{c})+8{\mathit{c}}^2(\mathit{S}-\mathit{r}.\mathit{c})$$$$$$$$\mathit{i}\mathit{t}\mathit{h}\mathit{i}\mathit{n}\mathit{k}\mathit{t}\mathit{h}\mathit{i}\mathit{s}\mathit{i}\mathit{s}\mathit{m}\mathit{a}\mathit{k}\mathit{e}\mathit{n}\mathit{o}\mathit{s}\mathit{e}\mathit{n}\mathit{s}\mathit{e}...$$

Commented by Shrinava last updated on 11/Aug/22

$$\mathrm{Dear}\mathrm{Sir},\mathrm{if}\mathrm{possible}\mathrm{could}\mathrm{you}\mathrm{please}$$$$\mathrm{give}\mathrm{the}\mathrm{complete}\mathrm{solution}$$

Commented by behi834171 last updated on 11/Aug/22

$$oksir.nowhavenotime.$$$$but,whatyoumeantfrom:\mathit{\mu }\left(\mathit{A}\right)?$$

Commented by Shrinava last updated on 12/Aug/22

$$\mathrm{Dear}\mathrm{Sir},\mathrm{was}\mathrm{the}\mathrm{end}\mathrm{result}\mathrm{the}$$$$\mathrm{answer}?$$

Commented by Shrinava last updated on 12/Aug/22

$$\mathrm{Dear}\mathrm{Sir},\mu \left(\mathrm{A}\right)\mathrm{in}\mathrm{radians}$$

Commented by Tawa11 last updated on 12/Aug/22

$$\mathrm{Weldone}\mathrm{sir}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com