Question Number 174779 by mnjuly1970 last updated on 10/Aug/22
Commented by behi834171 last updated on 10/Aug/22
$$\boldsymbol{{sir}}!\:\:\boldsymbol{{your}}\:\boldsymbol{{used}}\:\boldsymbol{{fonts}}\:\boldsymbol{{are}}\:\boldsymbol{{too}}\:\boldsymbol{{small}}. \\ $$$$\boldsymbol{{please}}\:\boldsymbol{{be}}\:\boldsymbol{{friend}}\:\boldsymbol{{with}}\:\boldsymbol{{my}}\:\boldsymbol{{eyes}}\:\boldsymbol{{and}} \\ $$$$\boldsymbol{{use}}\:\boldsymbol{{larger}}\:\boldsymbol{{fonts}}.\boldsymbol{{thanks}}\:\boldsymbol{{in}}\:\boldsymbol{{advance}}. \\ $$
Commented by mnjuly1970 last updated on 11/Aug/22
$$\:\:\:{yes}\:\:{sir}\:...{thanks}\:{alot}... \\ $$
Answered by behi834171 last updated on 10/Aug/22
![cos(A/2)=(√((p(p−a))/(bc))) and so on Πcos(A/2)=((p(√(p(p−a)(p−b)(p−c))))/(abc)) = =((p.S)/(abc))=(p/(4R))=((a+b+c)/(8R))=((2RΣsinA)/(8R))≤ ≤(1/4)×3×((√3)/2)=((3(√3))/8) .■ [note: ΣsinA≤3×sin(𝛑/3)=((3(√3))/2) ]](Q174780.png)
$$\boldsymbol{{cos}}\frac{\boldsymbol{{A}}}{\mathrm{2}}=\sqrt{\frac{\boldsymbol{{p}}\left(\boldsymbol{{p}}−\boldsymbol{{a}}\right)}{\boldsymbol{{bc}}}}\:\:\boldsymbol{{and}}\:\boldsymbol{{so}}\:\boldsymbol{{on}} \\ $$$$\Pi\boldsymbol{{cos}}\frac{\boldsymbol{{A}}}{\mathrm{2}}=\frac{\boldsymbol{{p}}\sqrt{\boldsymbol{{p}}\left(\boldsymbol{{p}}−\boldsymbol{{a}}\right)\left(\boldsymbol{{p}}−\boldsymbol{{b}}\right)\left(\boldsymbol{{p}}−\boldsymbol{{c}}\right)}}{\boldsymbol{{abc}}}\:= \\ $$$$=\frac{\boldsymbol{{p}}.\boldsymbol{{S}}}{\boldsymbol{{abc}}}=\frac{\boldsymbol{{p}}}{\mathrm{4}\boldsymbol{{R}}}=\frac{\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}}{\mathrm{8}\boldsymbol{{R}}}=\frac{\mathrm{2}\boldsymbol{{R}}\Sigma\boldsymbol{{sinA}}}{\mathrm{8}\boldsymbol{{R}}}\leqslant \\ $$$$\leqslant\frac{\mathrm{1}}{\mathrm{4}}×\mathrm{3}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}}\:\:\:\:.\blacksquare \\ $$$$\left[\boldsymbol{{note}}:\:\Sigma\boldsymbol{{sinA}}\leqslant\mathrm{3}×\boldsymbol{{sin}}\frac{\boldsymbol{\pi}}{\mathrm{3}}=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:\:\right] \\ $$