Question Number 180458 by mnjuly1970 last updated on 12/Nov/22
Answered by mr W last updated on 13/Nov/22
$${c}^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} +\mathrm{2}×\mathrm{1}×\sqrt{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{4}+\sqrt{\mathrm{3}} \\ $$$$\frac{\mathrm{sin}\:\alpha}{\:\sqrt{\mathrm{3}}}=\frac{\mathrm{sin}\:\mathrm{120}°}{\:\sqrt{\mathrm{4}+\sqrt{\mathrm{3}}}} \\ $$$$\mathrm{sin}\:\alpha=\sqrt{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}+\sqrt{\mathrm{3}}}}=\frac{\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{4}+\sqrt{\mathrm{3}}}} \\ $$$$\mathrm{cos}\:\left({x}+\alpha\right)=\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{4}+\sqrt{\mathrm{3}}}} \\ $$$${x}=\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{4}+\sqrt{\mathrm{3}}}}−\alpha \\ $$$${x}=\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{4}+\sqrt{\mathrm{3}}}}−\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{4}+\sqrt{\mathrm{3}}}}=\mathrm{15}° \\ $$
Commented by mr W last updated on 13/Nov/22
Answered by a.lgnaoui last updated on 12/Nov/22
![△BCE BE=BC+CE−2BC×CEcos (120) BE^2 =4+(√3) ⇒ BE=(√(4+(√3))) △ABE AB=(√(BE^2 −AE^2 )) =(√(2+(√3))) △BHC BC^2 =HC^2 +HB^2 = (AE−DE)^2 +(AB−CD)^2 3=((√2) −cos x)^2 +((√(2+(√3) )) −sin x)^2 3=5+(√3)−2(√2) cos x −2(√(2+(√3) )) sin x 0=2+(√3) −2((√2)cos x+(√(2+(√3))) sin x) 2((√2)cos x+((√((2+(√3))(1−cos^2 x)))=2+(√3) posons z=cosx (2+(√3) )(1−cos^2 x)=(((2+(√3))/2) −(√2) cos x)^2 2 −(2+(√3) )z^2 =(7/4)+2z^2 −(√2) (2+(√3) )z 2−[(4+(√3))]z^2 +(√2) (2+(√3) )z− (7/4) =0 z^2 −(((√2) (2+(√3)))/(4+(√3)))z−(1/(4(4+(√(3))) ))=0 △=((2(2+(√3) )^2 )/((4+(√3) )^2 )) +(1/((4+(√3) ))) z=((18+9(√3) )/((4+(√3))^2 ))=((9(2+(√3)))/((4+(√3) )^2 )) z=(1/2)[(((√2) (2+(√3)))/((4+(√3))))±((3(√(2+(√3))))/(4+(√3)))] x=((((√(2+(√3))) )[3±(√2) ((√(2+(√3))) )])/(2(4+(√3) ))) =(((√(4 +2(√3) )) ±3(√(2+(√3))) )/(8+2(√3))) x=cos^(−1) (z) x_1 =0,7323541177 x=74,75° x_2 =0,2630949343 x=43° x1:rejete ( ∡CDE >90) donc unique solution est x=∡CED=42,92≅43°](Q180505.png)
$$\bigtriangleup\mathrm{BCE}\:\:\:\:\mathrm{BE}=\mathrm{BC}+\mathrm{CE}−\mathrm{2BC}×\mathrm{CEcos}\:\left(\mathrm{120}\right) \\ $$$$\mathrm{BE}^{\mathrm{2}} =\mathrm{4}+\sqrt{\mathrm{3}}\:\:\:\:\:\:\:\Rightarrow\:\:\mathrm{BE}=\sqrt{\mathrm{4}+\sqrt{\mathrm{3}}} \\ $$$$\bigtriangleup\mathrm{ABE}\:\:\:\:\:\:\mathrm{AB}=\sqrt{\mathrm{BE}^{\mathrm{2}} −\mathrm{AE}^{\mathrm{2}} }\:=\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$$\bigtriangleup\mathrm{BHC}\:\:\:\:\mathrm{BC}^{\mathrm{2}} =\mathrm{HC}^{\mathrm{2}} +\mathrm{HB}^{\mathrm{2}} \:\:=\:\left(\mathrm{AE}−\mathrm{DE}\right)^{\mathrm{2}} +\left(\mathrm{AB}−\mathrm{CD}\right)^{\mathrm{2}} \\ $$$$\mathrm{3}=\left(\sqrt{\mathrm{2}}\:−\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}\:}\:−\mathrm{sin}\:\mathrm{x}\right)^{\mathrm{2}} \\ $$$$\mathrm{3}=\mathrm{5}+\sqrt{\mathrm{3}}−\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{cos}\:\mathrm{x}\:−\mathrm{2}\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}\:}\:\mathrm{sin}\:\mathrm{x} \\ $$$$\mathrm{0}=\mathrm{2}+\sqrt{\mathrm{3}}\:−\mathrm{2}\left(\sqrt{\mathrm{2}}\mathrm{cos}\:\mathrm{x}+\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:\mathrm{sin}\:\mathrm{x}\right) \\ $$$$\mathrm{2}\left(\sqrt{\mathrm{2}}\mathrm{cos}\:\mathrm{x}+\left(\sqrt{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\right.}\right)=\mathrm{2}+\sqrt{\mathrm{3}}\right. \\ $$$$\mathrm{posons}\:\:\mathrm{z}=\mathrm{cosx}\:\: \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right)\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\right)=\left(\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:−\sqrt{\mathrm{2}}\:\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} \\ $$$$\:\mathrm{2}\:−\left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right)\mathrm{z}^{\mathrm{2}} =\frac{\mathrm{7}}{\mathrm{4}}+\mathrm{2z}^{\mathrm{2}} −\sqrt{\mathrm{2}}\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right)\mathrm{z} \\ $$$$\mathrm{2}−\left[\left(\mathrm{4}+\sqrt{\mathrm{3}}\right)\right]\mathrm{z}^{\mathrm{2}} +\sqrt{\mathrm{2}}\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right)\mathrm{z}−\:\frac{\mathrm{7}}{\mathrm{4}}\:\:\:\:=\mathrm{0} \\ $$$$\mathrm{z}^{\mathrm{2}} −\frac{\sqrt{\mathrm{2}}\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)}{\mathrm{4}+\sqrt{\mathrm{3}}}\mathrm{z}−\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{4}+\sqrt{\left.\mathrm{3}\right)}\:\right.}=\mathrm{0} \\ $$$$\bigtriangleup=\frac{\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right)^{\mathrm{2}} }{\left(\mathrm{4}+\sqrt{\mathrm{3}}\:\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\left(\mathrm{4}+\sqrt{\mathrm{3}}\:\right)} \\ $$$$\: \\ $$$$\mathrm{z}=\frac{\mathrm{18}+\mathrm{9}\sqrt{\mathrm{3}}\:}{\left(\mathrm{4}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }=\frac{\mathrm{9}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)}{\left(\mathrm{4}+\sqrt{\mathrm{3}}\:\right)^{\mathrm{2}} } \\ $$$$\mathrm{z}=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\sqrt{\mathrm{2}}\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)}{\left(\mathrm{4}+\sqrt{\mathrm{3}}\right)}\pm\frac{\mathrm{3}\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}{\mathrm{4}+\sqrt{\mathrm{3}}}\right] \\ $$$$\: \\ $$$$\mathrm{x}=\frac{\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:\right)\left[\mathrm{3}\pm\sqrt{\mathrm{2}}\:\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:\right)\right]}{\mathrm{2}\left(\mathrm{4}+\sqrt{\mathrm{3}}\:\right)} \\ $$$$=\frac{\sqrt{\mathrm{4}\:+\mathrm{2}\sqrt{\mathrm{3}}\:}\:\:\pm\mathrm{3}\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:}{\mathrm{8}+\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$$\mathrm{x}=\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{z}\right) \\ $$$$\mathrm{x}_{\mathrm{1}} =\mathrm{0},\mathrm{7323541177}\:\:\:\:\:\:\:\mathrm{x}=\mathrm{74},\mathrm{75}° \\ $$$$\mathrm{x}_{\mathrm{2}} =\mathrm{0},\mathrm{2630949343}\:\:\:\:\mathrm{x}=\mathrm{43}° \\ $$$$\mathrm{x1}:\mathrm{rejete}\:\:\:\:\:\left(\:\measuredangle\mathrm{CDE}\:>\mathrm{90}\right) \\ $$$${donc}\:{unique}\:{solution}\:{est} \\ $$$$ \\ $$$$\:\mathrm{x}=\measuredangle\mathrm{CED}=\mathrm{42},\mathrm{92}\cong\mathrm{43}° \\ $$$$ \\ $$
Commented by a.lgnaoui last updated on 12/Nov/22
$${If}\:{there}\:{is}\:{no}\:{error}\:{in}\: \\ $$$${the}\:{calcul}. \\ $$
Commented by a.lgnaoui last updated on 12/Nov/22
