Question Number 211294 by Jubr last updated on 05/Sep/24
Commented by a.lgnaoui last updated on 05/Sep/24
Commented by Ghisom last updated on 05/Sep/24
$$\mathrm{just}\:\mathrm{saying}:\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{convex} \\ $$$${x}=\mathrm{arctan}\:\frac{\mathrm{3}\left(\mathrm{7}\sqrt{\mathrm{3}}−\mathrm{4}\sqrt{\mathrm{2}}\right)}{\mathrm{23}}\:\approx\mathrm{40}°\mathrm{9}'\mathrm{2}'' \\ $$$$\mathrm{let}\:{A}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\wedge\:{B}=\begin{pmatrix}{\mathrm{1}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\Rightarrow \\ $$$${C}=\begin{pmatrix}{\frac{\mathrm{83}+\mathrm{18}\sqrt{\mathrm{6}}}{\mathrm{43}}}\\{−\frac{\mathrm{6}\left(\mathrm{5}\sqrt{\mathrm{3}}−\mathrm{4}\sqrt{\mathrm{2}}\right)}{\mathrm{43}}}\end{pmatrix}\:\:\approx\begin{pmatrix}{\mathrm{2}.\mathrm{95560}}\\{−.\mathrm{419079}}\end{pmatrix} \\ $$$${D}=\begin{pmatrix}{\frac{\mathrm{2}\left(\mathrm{7}+\mathrm{9}\sqrt{\mathrm{6}}\right)}{\mathrm{19}}}\\{\frac{\mathrm{6}\left(\mathrm{7}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}\right)}{\mathrm{19}}}\end{pmatrix}\:\:\approx\begin{pmatrix}{\mathrm{3}.\mathrm{05741}}\\{\mathrm{2}.\mathrm{57919}}\end{pmatrix} \\ $$
Commented by Jubr last updated on 05/Sep/24
$${Thanks}\:{sirs} \\ $$
Answered by A5T last updated on 05/Sep/24
![4^2 +1^2 −2×4×1cosx=3^2 +2^2 −2×3×2cos(120−x) ⇒1=2cosx−3cos(120−x)⇒cosx=((7+9(√6))/(38)) ⇒sinx=((21(√2)−3(√3))/(38))⇒sin(120−x)=(((√3)+12(√2))/(19)) ⇒[ABCD]=((21(√2)−3(√3))/(19))+((3(√3)+36(√2))/(19))=((57(√2))/(19))=3(√2)](https://www.tinkutara.com/question/Q211297.png)
$$\mathrm{4}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} −\mathrm{2}×\mathrm{4}×\mathrm{1}{cosx}=\mathrm{3}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} −\mathrm{2}×\mathrm{3}×\mathrm{2}{cos}\left(\mathrm{120}−{x}\right) \\ $$$$\Rightarrow\mathrm{1}=\mathrm{2}{cosx}−\mathrm{3}{cos}\left(\mathrm{120}−{x}\right)\Rightarrow{cosx}=\frac{\mathrm{7}+\mathrm{9}\sqrt{\mathrm{6}}}{\mathrm{38}} \\ $$$$\Rightarrow{sinx}=\frac{\mathrm{21}\sqrt{\mathrm{2}}−\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{38}}\Rightarrow{sin}\left(\mathrm{120}−{x}\right)=\frac{\sqrt{\mathrm{3}}+\mathrm{12}\sqrt{\mathrm{2}}}{\mathrm{19}} \\ $$$$\Rightarrow\left[{ABCD}\right]=\frac{\mathrm{21}\sqrt{\mathrm{2}}−\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{19}}+\frac{\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{36}\sqrt{\mathrm{2}}}{\mathrm{19}}=\frac{\mathrm{57}\sqrt{\mathrm{2}}}{\mathrm{19}}=\mathrm{3}\sqrt{\mathrm{2}} \\ $$
Commented by Jubr last updated on 05/Sep/24
$${Thanks}\:{sir} \\ $$
Answered by a.lgnaoui last updated on 05/Sep/24
![PHASE 1 (Shema 1) Area(ABCD)=Area[(ABD)+(BCD)] 1•Area(ABD)=(1/2)4.1.sin x Area(ABD)=2sin x 2•Area(BCD)=(1/2).3.2sin y AArea(BCD)=3sin y ⇒Area(ABCD)=2sin x+3sin y y=120−x Area(ABCD)=2sin x+3cos (30−x) =2sin x+(3/2)((√3)cos x+sin x) Area(ABCD)=(7/2)sin x+((3(√3))/2)cos x Or x+y+∡B+∡D=360 PHASE 2 BD cote commun BD^2 =17−8cos x=13−12cos y soit 17−8cosx=13+12sin (30−x) 4 1=3sin (30−x)+2cos x 3((1/2)cos x−((√3)/2)sin x)+2cos x=1 (7/2)cos x−((3(√3))/2)sin x=1 7(√(1−sin^2 x)) =3(√3)sin x+1(2) 49(1−sin^2 x)=27sin^2 x+1+6(√3) sin x 38sin^2 x+3(√3) sin x−24=0 △=27+96.38=[35(√3) ]^2 sin x=((−3(√3) ±35(√3))/(76))=((8(√3))/(19)) cos x=(√(1−((64×3)/(19^2 )))) =((13)/(19)) PHASE(1)⇒ Area(ABCD)=(7/2)sin x+3((√3)/2)cos x =(((56+39)(√3))/(38)) (After rectificat error in calcul the value is) Area(ABCD) =2(√3)](https://www.tinkutara.com/question/Q211298.png)
$$\boldsymbol{\mathrm{PHASE}}\:\mathrm{1}\:\:\:\left(\boldsymbol{\mathrm{Shema}}\:\mathrm{1}\right) \\ $$$$\mathrm{Area}\left(\mathrm{ABCD}\right)=\mathrm{Area}\left[\left(\mathrm{ABD}\right)+\left(\mathrm{BCD}\right)\right] \\ $$$$\mathrm{1}\bullet\mathrm{Area}\left(\mathrm{ABD}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{4}.\mathrm{1}.\mathrm{sin}\:\boldsymbol{\mathrm{x}} \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{Area}}\left(\boldsymbol{\mathrm{ABD}}\right)=\mathrm{2sin}\:\boldsymbol{\mathrm{x}} \\ $$$$\mathrm{2}\bullet\mathrm{Area}\left(\mathrm{BCD}\right)=\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{3}.\mathrm{2sin}\:\boldsymbol{\mathrm{y}} \\ $$$$\:\:\:\:\:\:\boldsymbol{\mathrm{A}}\mathrm{A}\boldsymbol{\mathrm{rea}}\left(\boldsymbol{\mathrm{BCD}}\right)=\mathrm{3sin}\:\boldsymbol{\mathrm{y}} \\ $$$$\Rightarrow\mathrm{Area}\left(\mathrm{ABCD}\right)=\mathrm{2sin}\:\boldsymbol{\mathrm{x}}+\mathrm{3sin}\:\boldsymbol{\mathrm{y}} \\ $$$$\:\:\boldsymbol{\mathrm{y}}=\mathrm{120}−\boldsymbol{\mathrm{x}} \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{Area}}\left(\boldsymbol{\mathrm{ABCD}}\right)=\mathrm{2sin}\:\boldsymbol{\mathrm{x}}+\mathrm{3cos}\:\:\left(\mathrm{30}−\boldsymbol{\mathrm{x}}\right) \\ $$$$\:\:\:\:=\mathrm{2sin}\:\boldsymbol{\mathrm{x}}+\frac{\mathrm{3}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}\mathrm{cos}\:\boldsymbol{\mathrm{x}}+\mathrm{sin}\:\boldsymbol{\mathrm{x}}\right) \\ $$$$\:\:\:\boldsymbol{\mathrm{Area}}\left(\boldsymbol{\mathrm{ABCD}}\right)=\frac{\mathrm{7}}{\mathrm{2}}\mathrm{sin}\:\boldsymbol{\mathrm{x}}+\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:\boldsymbol{\mathrm{x}} \\ $$$$\boldsymbol{\mathrm{Or}}\:\:\:\:\:\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}+\measuredangle\boldsymbol{\mathrm{B}}+\measuredangle\boldsymbol{\mathrm{D}}=\mathrm{360} \\ $$$$\:\:\boldsymbol{\mathrm{PHASE}}\:\mathrm{2} \\ $$$$\:\:\boldsymbol{\mathrm{BD}}\:\mathrm{cote}\:\mathrm{commun} \\ $$$$\:\mathrm{BD}^{\mathrm{2}} =\mathrm{17}−\mathrm{8cos}\:\boldsymbol{\mathrm{x}}=\mathrm{13}−\mathrm{12cos}\:\boldsymbol{\mathrm{y}} \\ $$$$\:\:\:\boldsymbol{\mathrm{soit}}\:\:\:\mathrm{17}−\mathrm{8cos}\boldsymbol{\mathrm{x}}=\mathrm{13}+\mathrm{12sin}\:\left(\mathrm{30}−\boldsymbol{\mathrm{x}}\right) \\ $$$$\mathrm{4} \\ $$$$\:\:\:\:\:\mathrm{1}=\mathrm{3sin}\:\left(\mathrm{30}−\boldsymbol{\mathrm{x}}\right)+\mathrm{2cos}\:\boldsymbol{\mathrm{x}} \\ $$$$\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\boldsymbol{\mathrm{x}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}\:\boldsymbol{\mathrm{x}}\right)+\mathrm{2cos}\:\boldsymbol{\mathrm{x}}=\mathrm{1} \\ $$$$\:\:\:\frac{\mathrm{7}}{\mathrm{2}}\mathrm{cos}\:\boldsymbol{\mathrm{x}}−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sin}\:\boldsymbol{\mathrm{x}}=\mathrm{1} \\ $$$$\:\:\mathrm{7}\sqrt{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \boldsymbol{\mathrm{x}}}\:=\mathrm{3}\sqrt{\mathrm{3}}\mathrm{sin}\:\boldsymbol{\mathrm{x}}+\mathrm{1}\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\mathrm{49}\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \boldsymbol{\mathrm{x}}\right)=\mathrm{27sin}\:^{\mathrm{2}} \boldsymbol{\mathrm{x}}+\mathrm{1}+\mathrm{6}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\boldsymbol{\mathrm{x}} \\ $$$$\:\:\:\:\mathrm{38sin}\:^{\mathrm{2}} \boldsymbol{\mathrm{x}}+\mathrm{3}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\boldsymbol{\mathrm{x}}−\mathrm{24}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\bigtriangleup=\mathrm{27}+\mathrm{96}.\mathrm{38}=\left[\mathrm{35}\sqrt{\mathrm{3}}\:\right]^{\mathrm{2}} \\ $$$$ \\ $$$$\:\:\:\mathrm{sin}\:\boldsymbol{\mathrm{x}}=\frac{−\mathrm{3}\sqrt{\mathrm{3}}\:\pm\mathrm{35}\sqrt{\mathrm{3}}}{\mathrm{76}}=\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{19}} \\ $$$$\:\:\:\mathrm{cos}\:\mathrm{x}=\sqrt{\mathrm{1}−\frac{\mathrm{64}×\mathrm{3}}{\mathrm{19}^{\mathrm{2}} }}\:=\frac{\mathrm{13}}{\mathrm{19}} \\ $$$$\mathrm{PHASE}\left(\mathrm{1}\right)\Rightarrow \\ $$$$ \\ $$$$\:\:\mathrm{Area}\left(\mathrm{ABCD}\right)=\frac{\mathrm{7}}{\mathrm{2}}\mathrm{sin}\:\boldsymbol{\mathrm{x}}+\mathrm{3}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:\boldsymbol{\mathrm{x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{56}+\mathrm{39}\right)\sqrt{\mathrm{3}}}{\mathrm{38}} \\ $$$$\left({After}\:\:{rectificat}\:{error}\:{in}\:{calcul}\:{the}\:{value}\:{is}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{Area}}\left(\boldsymbol{\mathrm{ABCD}}\right)\:\:\:=\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\:\:\: \\ $$$$\: \\ $$$$\: \\ $$
Commented by A5T last updated on 05/Sep/24
$$\frac{\mathrm{7}}{\mathrm{2}}{cosx}−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}{sinx}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{7}\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} {x}}=\mathrm{3}\sqrt{\mathrm{3}}{sinx}+\mathrm{2} \\ $$$$\Rightarrow\mathrm{49}\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right)=\mathrm{27}{sin}^{\mathrm{2}} {x}+\mathrm{4}+\mathrm{12}\sqrt{\mathrm{3}}{sinx} \\ $$
Commented by a.lgnaoui last updated on 05/Sep/24
$$\mathrm{ok}\:\mathrm{merci}. \\ $$
Commented by Jubr last updated on 05/Sep/24
$${Thanks}\:{sir} \\ $$