Question-179752

Question Number 179752 by Acem last updated on 01/Nov/22

Commented by Acem last updated on 01/Nov/22

$${P}.{s}.\:{It}'{s}\:{not}\:{right}\:{triangle} \\ $$

Commented by CElcedricjunior last updated on 02/Nov/22

calculons A d′apres le theoreme des sinus ((18)/(sin𝛔))=((30)/(sin𝛃))=(k/(sin𝛉))=((540k)/(2A))=2r =>((18)/(sin(𝛃/2)))=((30)/(2sin(𝛃/2)cos(𝛃/2)))=(k/(sin𝛉)) =>2cos(𝛃/2)=((10)/6)=(5/3) =>𝛃=2arcos((5/6)) =>𝛃=67.11=>𝛂=((67.11)/2)=33.55 =>((30)/(sin(2arccos((5/6)))))=(k/(sin(180−3arcos((5/6))))) k=((30sin(180−3arccos((5/6))))/(sin(2arccos((5/6))))) ⇔((18)/(sin(arccos((5/6)))))=((270×30sin(180−3arccos((5/6))))/(Asin(2arccos((5/6))))) ⇔A=((8100sin(180−3arccos((5/6))))/(36sin(arccos((5/6))))) formule brute .....................le celebre cedric junior........

$$\boldsymbol{\mathrm{calculons}}\:\boldsymbol{\mathrm{A}} \\ $$$$\boldsymbol{{d}}'\boldsymbol{{apres}}\:\boldsymbol{{le}}\:\boldsymbol{{theoreme}}\:\boldsymbol{{des}}\:\boldsymbol{{sinus}} \\ $$$$\frac{\mathrm{18}}{\boldsymbol{{sin}\sigma}}=\frac{\mathrm{30}}{\boldsymbol{{sin}\beta}}=\frac{\boldsymbol{{k}}}{\boldsymbol{{sin}\theta}}=\frac{\mathrm{540}\boldsymbol{{k}}}{\mathrm{2}\boldsymbol{{A}}}=\mathrm{2}{r} \\ $$$$=>\frac{\mathrm{18}}{{sin}\frac{\boldsymbol{\beta}}{\mathrm{2}}}=\frac{\mathrm{30}}{\mathrm{2}{sin}\frac{\boldsymbol{\beta}}{\mathrm{2}}\boldsymbol{{cos}}\frac{\boldsymbol{\beta}}{\mathrm{2}}}=\frac{{k}}{\boldsymbol{{sin}\theta}} \\ $$$$=>\mathrm{2}\boldsymbol{{cos}}\frac{\boldsymbol{\beta}}{\mathrm{2}}=\frac{\mathrm{10}}{\mathrm{6}}=\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$=>\boldsymbol{\beta}=\mathrm{2}\boldsymbol{{arcos}}\left(\frac{\mathrm{5}}{\mathrm{6}}\right) \\ $$$$=>\boldsymbol{\beta}=\mathrm{67}.\mathrm{11}=>\boldsymbol{\alpha}=\frac{\mathrm{67}.\mathrm{11}}{\mathrm{2}}=\mathrm{33}.\mathrm{55} \\ $$$$=>\frac{\mathrm{30}}{\boldsymbol{{sin}}\left(\mathrm{2}\boldsymbol{{arccos}}\left(\frac{\mathrm{5}}{\mathrm{6}}\right)\right)}=\frac{\boldsymbol{{k}}}{\boldsymbol{{sin}}\left(\mathrm{180}−\mathrm{3}\boldsymbol{{arcos}}\left(\frac{\mathrm{5}}{\mathrm{6}}\right)\right)} \\ $$$$\boldsymbol{{k}}=\frac{\mathrm{30}\boldsymbol{{sin}}\left(\mathrm{180}−\mathrm{3}\boldsymbol{{arccos}}\left(\frac{\mathrm{5}}{\mathrm{6}}\right)\right)}{\boldsymbol{{sin}}\left(\mathrm{2}\boldsymbol{{arccos}}\left(\frac{\mathrm{5}}{\mathrm{6}}\right)\right)} \\ $$$$\Leftrightarrow\frac{\mathrm{18}}{\boldsymbol{{sin}}\left(\boldsymbol{{arccos}}\left(\frac{\mathrm{5}}{\mathrm{6}}\right)\right)}=\frac{\mathrm{270}×\mathrm{30}\boldsymbol{{sin}}\left(\mathrm{180}−\mathrm{3}\boldsymbol{{arccos}}\left(\frac{\mathrm{5}}{\mathrm{6}}\right)\right)}{\boldsymbol{{Asin}}\left(\mathrm{2}\boldsymbol{{arccos}}\left(\frac{\mathrm{5}}{\mathrm{6}}\right)\right)} \\ $$$$\Leftrightarrow\boldsymbol{{A}}=\frac{\mathrm{8100}\boldsymbol{{sin}}\left(\mathrm{180}−\mathrm{3}\boldsymbol{{arccos}}\left(\frac{\mathrm{5}}{\mathrm{6}}\right)\right)}{\mathrm{36}\boldsymbol{{sin}}\left(\boldsymbol{{arccos}}\left(\frac{\mathrm{5}}{\mathrm{6}}\right)\right)} \\ $$$$\boldsymbol{{formule}}\:\boldsymbol{{brute}}\: \\ $$$$…………………{le}\:{celebre}\:\:{cedric}\:{junior}…….. \\ $$

Answered by Acem last updated on 02/Nov/22

$${What}\:{if}\:{we}\:{don}'{t}\:{know}\:{or}\:{forgot}\:{at}\:{least}\:{one}\:{of} \\ $$$$\:{the}\:{formulas}\:{that}\:{our}\:{friend}\:{AST}\:{has}\:{used}! \\ $$$$\:{or}\:{even}\:{hadn}'{t}\:{thought}\:{of}\:{his}\:{idea}\:{at}\:{solve}. \\ $$$$\:{We}'{ll}\:{go}\:{for}\:{a}\:{trick}\:{as}\:{follow}: \\ $$

Commented by Acem last updated on 02/Nov/22

Commented by Acem last updated on 02/Nov/22

The square height is equal: h^2 = a^2 − c^2 = b^2 − (c+a)^(2 ) ⇒ c= 7, h= (√(275))= 5 (√(11)) Area= (1/2) 32× 5 (√(11)) = 80 (√(11)) ≈ 265.33 un.^2

$$\:{The}\:{square}\:{height}\:{is}\:{equal}: \\ $$$$\:{h}^{\mathrm{2}} =\:{a}^{\mathrm{2}} −\:{c}^{\mathrm{2}} \:=\:{b}^{\mathrm{2}} −\:\left({c}+{a}\right)^{\mathrm{2}\:} \:\Rightarrow\:{c}=\:\mathrm{7},\:{h}=\:\sqrt{\mathrm{275}}=\:\mathrm{5}\:\sqrt{\mathrm{11}} \\ $$$$\:{Area}=\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{32}×\:\mathrm{5}\:\sqrt{\mathrm{11}}\:=\:\mathrm{80}\:\sqrt{\mathrm{11}}\:\approx\:\mathrm{265}.\mathrm{33}\:\:{un}.^{\mathrm{2}} \\ $$

Commented by manxsol last updated on 02/Nov/22