Question Number 174799 by Shrinava last updated on 11/Aug/22
Answered by behi834171 last updated on 12/Aug/22
![cot(A/2)=(√((p(p−a))/((p−b)(p−c)))) =(S/((p−b)(p−c)))= =((p(p−a) )/S)=((p−a)/r) and so on... Πcot(A/2)=(S^2 /(p.r^3 ))=(S/r^2 ) (lhs)^2 =(((p−a)^2 )/r^2 )+(((p−b)^2 )/r^2 )+4×((p−c)/r)+ +2×(((p−a)(p−b))/r^2 )+4×((((p−a))/r)+(((p−b))/r)).(√((p−c)/r))= =[(((p−a)^2 +(p−b)^2 +2(p−a)(p−b))/r^2 )]+ +((4(p−c))/r)+4×((2p−a−b)/r)(√((p−c)/r))= =(c^2 /r^2 )+4×(c/r)(√((p−c)/r))+4×((p−c)/r)= =((c/r)+2(√((p−c)/r)))^2 =(((c+2(√(S−r.c)))^2 )/r^2 ) (lhs)^4 =((c^2 +4(S−r.c)+4(√(S−r.c)))/r^4 ) (RHS)^4 =4^4 ×(S/r^2 )⇒ ⇒c^2 +4(S−r.c)+4(√(S−r.c))=256S.r^2 ⇒ ⇒16(S−r.c)=256^2 S^2 .r^4 +c^4 +16(S−r.c)^2 − −512Sr^2 c^2 −512×8S.r^2 .(S−r.c)+8c^2 (S−r.c) i think this is make no sense...](https://www.tinkutara.com/question/Q174817.png)
$${cot}\frac{{A}}{\mathrm{2}}=\sqrt{\frac{{p}\left({p}−{a}\right)}{\left({p}−{b}\right)\left({p}−{c}\right)}}\:\:=\frac{\boldsymbol{{S}}}{\left(\boldsymbol{{p}}−\boldsymbol{{b}}\right)\left(\boldsymbol{{p}}−\boldsymbol{{c}}\right)}= \\ $$$$=\frac{\boldsymbol{{p}}\left(\boldsymbol{{p}}−\boldsymbol{{a}}\right)\:}{\boldsymbol{{S}}}=\frac{\boldsymbol{{p}}−\boldsymbol{{a}}}{\boldsymbol{{r}}}\:\:\:\:\:{and}\:{so}\:{on}… \\ $$$$\Pi\boldsymbol{{cot}}\frac{\boldsymbol{{A}}}{\mathrm{2}}=\frac{\boldsymbol{{S}}^{\mathrm{2}} }{\boldsymbol{{p}}.\boldsymbol{{r}}^{\mathrm{3}} }=\frac{\boldsymbol{{S}}}{\boldsymbol{{r}}^{\mathrm{2}} } \\ $$$$\left(\boldsymbol{{lhs}}\right)^{\mathrm{2}} =\frac{\left(\boldsymbol{{p}}−\boldsymbol{{a}}\right)^{\mathrm{2}} }{\boldsymbol{{r}}^{\mathrm{2}} }+\frac{\left(\boldsymbol{{p}}−\boldsymbol{{b}}\right)^{\mathrm{2}} }{\boldsymbol{{r}}^{\mathrm{2}} }+\mathrm{4}×\frac{\boldsymbol{{p}}−\boldsymbol{{c}}}{\boldsymbol{{r}}}+ \\ $$$$+\mathrm{2}×\frac{\left(\boldsymbol{{p}}−\boldsymbol{{a}}\right)\left(\boldsymbol{{p}}−\boldsymbol{{b}}\right)}{\boldsymbol{{r}}^{\mathrm{2}} }+\mathrm{4}×\left(\frac{\left(\boldsymbol{{p}}−\boldsymbol{{a}}\right)}{\boldsymbol{{r}}}+\frac{\left(\boldsymbol{{p}}−\boldsymbol{{b}}\right)}{\boldsymbol{{r}}}\right).\sqrt{\frac{\boldsymbol{{p}}−\boldsymbol{{c}}}{\boldsymbol{{r}}}}= \\ $$$$=\left[\frac{\left(\boldsymbol{{p}}−\boldsymbol{{a}}\right)^{\mathrm{2}} +\left(\boldsymbol{{p}}−\boldsymbol{{b}}\right)^{\mathrm{2}} +\mathrm{2}\left(\boldsymbol{{p}}−\boldsymbol{{a}}\right)\left(\boldsymbol{{p}}−\boldsymbol{{b}}\right)}{\boldsymbol{{r}}^{\mathrm{2}} }\right]+ \\ $$$$+\frac{\mathrm{4}\left(\boldsymbol{{p}}−\boldsymbol{{c}}\right)}{\boldsymbol{{r}}}+\mathrm{4}×\frac{\mathrm{2}\boldsymbol{{p}}−\boldsymbol{{a}}−\boldsymbol{{b}}}{\boldsymbol{{r}}}\sqrt{\frac{\boldsymbol{{p}}−\boldsymbol{{c}}}{\boldsymbol{{r}}}}= \\ $$$$=\frac{\boldsymbol{{c}}^{\mathrm{2}} }{\boldsymbol{{r}}^{\mathrm{2}} }+\mathrm{4}×\frac{\boldsymbol{{c}}}{\boldsymbol{{r}}}\sqrt{\frac{\boldsymbol{{p}}−\boldsymbol{{c}}}{\boldsymbol{{r}}}}+\mathrm{4}×\frac{\boldsymbol{{p}}−\boldsymbol{{c}}}{\boldsymbol{{r}}}= \\ $$$$=\left(\frac{\boldsymbol{{c}}}{\boldsymbol{{r}}}+\mathrm{2}\sqrt{\frac{\boldsymbol{{p}}−\boldsymbol{{c}}}{\boldsymbol{{r}}}}\right)^{\mathrm{2}} =\frac{\left(\boldsymbol{{c}}+\mathrm{2}\sqrt{\boldsymbol{{S}}−\boldsymbol{{r}}.\boldsymbol{{c}}}\right)^{\mathrm{2}} }{\boldsymbol{{r}}^{\mathrm{2}} } \\ $$$$\left(\boldsymbol{{lhs}}\right)^{\mathrm{4}} =\frac{\boldsymbol{{c}}^{\mathrm{2}} +\mathrm{4}\left(\boldsymbol{{S}}−\boldsymbol{{r}}.\boldsymbol{{c}}\right)+\mathrm{4}\sqrt{\boldsymbol{{S}}−\boldsymbol{{r}}.\boldsymbol{{c}}}}{\boldsymbol{{r}}^{\mathrm{4}} } \\ $$$$\left(\boldsymbol{{RHS}}\right)^{\mathrm{4}} =\mathrm{4}^{\mathrm{4}} ×\frac{\boldsymbol{{S}}}{\boldsymbol{{r}}^{\mathrm{2}} }\Rightarrow \\ $$$$\Rightarrow\boldsymbol{{c}}^{\mathrm{2}} +\mathrm{4}\left(\boldsymbol{{S}}−\boldsymbol{{r}}.\boldsymbol{{c}}\right)+\mathrm{4}\sqrt{\boldsymbol{{S}}−\boldsymbol{{r}}.\boldsymbol{{c}}}=\mathrm{256}\boldsymbol{{S}}.\boldsymbol{{r}}^{\mathrm{2}} \Rightarrow \\ $$$$\Rightarrow\mathrm{16}\left(\boldsymbol{{S}}−\boldsymbol{{r}}.\boldsymbol{{c}}\right)=\mathrm{256}^{\mathrm{2}} \boldsymbol{{S}}^{\mathrm{2}} .\boldsymbol{{r}}^{\mathrm{4}} +\boldsymbol{{c}}^{\mathrm{4}} +\mathrm{16}\left(\boldsymbol{{S}}−\boldsymbol{{r}}.\boldsymbol{{c}}\right)^{\mathrm{2}} − \\ $$$$−\mathrm{512}\boldsymbol{{Sr}}^{\mathrm{2}} \boldsymbol{{c}}^{\mathrm{2}} −\mathrm{512}×\mathrm{8}\boldsymbol{{S}}.\boldsymbol{{r}}^{\mathrm{2}} .\left(\boldsymbol{{S}}−\boldsymbol{{r}}.\boldsymbol{{c}}\right)+\mathrm{8}\boldsymbol{{c}}^{\mathrm{2}} \left(\boldsymbol{{S}}−\boldsymbol{{r}}.\boldsymbol{{c}}\right) \\ $$$$ \\ $$$$\boldsymbol{{i}}\:\boldsymbol{{think}}\:\boldsymbol{{this}}\:\boldsymbol{{is}}\:\boldsymbol{{make}}\:\boldsymbol{{no}}\:\boldsymbol{{sense}}… \\ $$
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
$${ok}\:{sir}.\:{now}\:{have}\:{no}\:{time}. \\ $$$${but},{what}\:{you}\:{meant}\:{from}:\boldsymbol{\mu}\left(\boldsymbol{{A}}\right)? \\ $$
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
$$\mathrm{Weldone}\:\mathrm{sir} \\ $$