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Question Number 195043 by necx122 last updated on 22/Jul/23
Commented by necx122 last updated on 22/Jul/23
Please, I meed help with this. Itβ€²s quite tricky for me to tackle.
$${Please},\:{I}\:{meed}\:{help}\:{with}\:{this}.\:{It}'{s}\:{quite} \\ $$$${tricky}\:{for}\:{me}\:{to}\:{tackle}. \\ $$
Answered by a.lgnaoui last updated on 23/Jul/23
β€’Cote AB=a+b { ((sin 𝛂=(2/a)β‡’ a=(2/(sin 𝛂)))),((cos 𝛂=(2/b)β‡’b=(2/(cos 𝛂)))) :} a+b=2(((sin 𝛂+cos Ξ±)/(sin 𝛂cos 𝛂))) β€’Cote AC=c+d+e { ((sin (90βˆ’π›‚)=(3/c)β‡’ c=(3/(cos 𝛂)))),((sin (90βˆ’π›‚)=(2/d)=(1/e))) :} β‡’c=(3/(cos 𝛂)) ;d=(2/(cos 𝛂)) ; e=(d/2) alors : AC=c+d+d=(6/(cos 𝛂)) AB=AC (quadrilatere=care) β‡’(6/(cos 𝛂))=((2(sin 𝛂+cos 𝛂)/(sin 𝛂cos 𝛂)) 4sin 𝛂cos 𝛂=sin 𝛂cos 𝛂+cos^2 𝛂 4=1+((cos 𝛂)/(sin 𝛂)) β‡’ tan 𝛂=(1/3) 𝛂= 18,43 donc AB=AC=(6/(cos 𝛂))=6(√(1+tan^2 𝛂)) cote du care =6(√(1+(1/9))) =2(√(10 )) β‡’ Aire du Care= 40
$$\:\bullet\boldsymbol{\mathrm{C}}\mathrm{ote}\:\boldsymbol{\mathrm{AB}}=\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}} \\ $$$$\:\begin{cases}{\mathrm{sin}\:\boldsymbol{\alpha}=\frac{\mathrm{2}}{\mathrm{a}}\Rightarrow\:\:\boldsymbol{\mathrm{a}}=\frac{\mathrm{2}}{\mathrm{sin}\:\boldsymbol{\alpha}}}\\{\mathrm{cos}\:\boldsymbol{\alpha}=\frac{\mathrm{2}}{\mathrm{b}}\Rightarrow\boldsymbol{\mathrm{b}}=\frac{\mathrm{2}}{\mathrm{cos}\:\boldsymbol{\alpha}}}\end{cases} \\ $$$$\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}=\mathrm{2}\left(\frac{\mathrm{sin}\:\boldsymbol{\alpha}+\mathrm{cos}\:\alpha}{\mathrm{sin}\:\boldsymbol{\alpha}\mathrm{cos}\:\boldsymbol{\alpha}}\right) \\ $$$$ \\ $$$$\bullet\boldsymbol{\mathrm{Cote}}\:\:\boldsymbol{\mathrm{AC}}=\boldsymbol{\mathrm{c}}+\boldsymbol{\mathrm{d}}+\boldsymbol{\mathrm{e}} \\ $$$$\:\:\begin{cases}{\mathrm{sin}\:\left(\mathrm{90}βˆ’\boldsymbol{\alpha}\right)=\frac{\mathrm{3}}{\mathrm{c}}\Rightarrow\:\:\boldsymbol{\mathrm{c}}=\frac{\mathrm{3}}{\mathrm{cos}\:\boldsymbol{\alpha}}}\\{\mathrm{sin}\:\left(\mathrm{90}βˆ’\boldsymbol{\alpha}\right)=\frac{\mathrm{2}}{\boldsymbol{\mathrm{d}}}=\frac{\mathrm{1}}{\boldsymbol{\mathrm{e}}}}\end{cases} \\ $$$$\Rightarrow\boldsymbol{\mathrm{c}}=\frac{\mathrm{3}}{\mathrm{cos}\:\boldsymbol{\alpha}}\:\:;\boldsymbol{\mathrm{d}}=\frac{\mathrm{2}}{\mathrm{cos}\:\boldsymbol{\alpha}}\:;\:\:\boldsymbol{\mathrm{e}}=\frac{\boldsymbol{\mathrm{d}}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{alors}}\:\::\:\:\boldsymbol{\mathrm{AC}}=\boldsymbol{\mathrm{c}}+\boldsymbol{\mathrm{d}}+\boldsymbol{\mathrm{d}}=\frac{\mathrm{6}}{\mathrm{cos}\:\boldsymbol{\alpha}} \\ $$$$\:\:\boldsymbol{\mathrm{AB}}=\boldsymbol{\mathrm{AC}}\:\:\:\left(\boldsymbol{\mathrm{quadrilatere}}=\boldsymbol{\mathrm{care}}\right) \\ $$$$\Rightarrow\frac{\mathrm{6}}{\mathrm{cos}\:\boldsymbol{\alpha}}=\frac{\mathrm{2}\left(\mathrm{sin}\:\boldsymbol{\alpha}+\mathrm{cos}\:\boldsymbol{\alpha}\right.}{\mathrm{sin}\:\boldsymbol{\alpha}\mathrm{cos}\:\boldsymbol{\alpha}}\:\:\: \\ $$$$\:\:\mathrm{4sin}\:\boldsymbol{\alpha}\mathrm{cos}\:\boldsymbol{\alpha}=\mathrm{sin}\:\boldsymbol{\alpha}\mathrm{cos}\:\boldsymbol{\alpha}+\mathrm{cos}\:^{\mathrm{2}} \boldsymbol{\alpha} \\ $$$$\mathrm{4}=\mathrm{1}+\frac{\mathrm{cos}\:\boldsymbol{\alpha}}{\mathrm{sin}\:\boldsymbol{\alpha}}\:\:\:\Rightarrow\:\:\:\mathrm{tan}\:\boldsymbol{\alpha}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\:\:\:\boldsymbol{\alpha}=\:\:\mathrm{18},\mathrm{43} \\ $$$$\boldsymbol{\mathrm{donc}}\:\:\boldsymbol{\mathrm{AB}}=\boldsymbol{\mathrm{AC}}=\frac{\mathrm{6}}{\mathrm{cos}\:\boldsymbol{\alpha}}=\mathrm{6}\sqrt{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \boldsymbol{\alpha}} \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{cote}}\:\boldsymbol{\mathrm{du}}\:\boldsymbol{\mathrm{care}}\:\:\:\:\:=\mathrm{6}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{9}}}\:=\mathrm{2}\sqrt{\mathrm{10}\:}\: \\ $$$$ \\ $$$$\:\:\:\Rightarrow\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{Aire}}\:\boldsymbol{\mathrm{du}}\:\boldsymbol{\mathrm{Care}}=\:\mathrm{40} \\ $$$$ \\ $$
Commented by a.lgnaoui last updated on 23/Jul/23
Commented by JDamian last updated on 23/Jul/23
I get 45 without trigonometric expressions
$$\mathrm{I}\:\mathrm{get}\:\mathrm{45}\:\mathrm{without}\:\mathrm{trigonometric}\:\mathrm{expressions} \\ $$
Commented by JDamian last updated on 23/Jul/23
(6/(cos 𝛂))=((2(sin 𝛂+cos 𝛂))/(sin 𝛂cos 𝛂)) 3=((sin 𝛂+cos 𝛂)/(sin 𝛂))=1+((cos 𝛂)/(sin 𝛂)) 2=((cos 𝛂)/(sin 𝛂)) tan 𝛂=(1/2) donc AB=AC=6(√(1+tan^2 𝛂))= =6(√(1+((1/2))^2 ))=6(√(5/4))=3(√5)
$$\frac{\mathrm{6}}{\cancel{\mathrm{cos}\:\boldsymbol{\alpha}}}=\frac{\mathrm{2}\left(\mathrm{sin}\:\boldsymbol{\alpha}+\mathrm{cos}\:\boldsymbol{\alpha}\right)}{\mathrm{sin}\:\boldsymbol{\alpha}\cancel{\mathrm{cos}\:\boldsymbol{\alpha}}} \\ $$$$\mathrm{3}=\frac{\mathrm{sin}\:\boldsymbol{\alpha}+\mathrm{cos}\:\boldsymbol{\alpha}}{\mathrm{sin}\:\boldsymbol{\alpha}}=\mathrm{1}+\frac{\mathrm{cos}\:\boldsymbol{\alpha}}{\mathrm{sin}\:\boldsymbol{\alpha}} \\ $$$$\mathrm{2}=\frac{\mathrm{cos}\:\boldsymbol{\alpha}}{\mathrm{sin}\:\boldsymbol{\alpha}} \\ $$$$\mathrm{tan}\:\boldsymbol{\alpha}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$$\boldsymbol{\mathrm{donc}}\:\:\boldsymbol{\mathrm{AB}}=\boldsymbol{\mathrm{AC}}=\mathrm{6}\sqrt{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \boldsymbol{\alpha}}= \\ $$$$=\mathrm{6}\sqrt{\mathrm{1}+\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }=\mathrm{6}\sqrt{\frac{\mathrm{5}}{\mathrm{4}}}=\mathrm{3}\sqrt{\mathrm{5}} \\ $$
Commented by a.lgnaoui last updated on 23/Jul/23
correction(erreur calcul) 3=1+((cos Ξ±)/(sin Ξ±)) β‡’tan Ξ±=(1/2) Cote=(6/(cos Ξ±))=6(√(1+tan^2 Ξ±)) =3(√5) Aire=45
$$\mathrm{correction}\left(\mathrm{erreur}\:\mathrm{calcul}\right) \\ $$$$\mathrm{3}=\mathrm{1}+\frac{\mathrm{cos}\:\alpha}{\mathrm{sin}\:\alpha}\:\Rightarrow\mathrm{tan}\:\alpha=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{Cote}=\frac{\mathrm{6}}{\mathrm{cos}\:\alpha}=\mathrm{6}\sqrt{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \alpha}\:=\mathrm{3}\sqrt{\mathrm{5}} \\ $$$$\mathrm{Aire}=\mathrm{45} \\ $$$$ \\ $$
Commented by necx122 last updated on 23/Jul/23
Also clear.
$${Also}\:{clear}. \\ $$
Answered by mr W last updated on 23/Jul/23
Commented by mr W last updated on 23/Jul/23
tan ΞΈ=(2/(3βˆ’1))=2 (6/a)=sin ΞΈ=(2/( (√(2^2 +1^2 ))))=(2/( (√5))) β‡’a=3(√5) area of square =a^2 =(3(√5))^2 =45
$$\mathrm{tan}\:\theta=\frac{\mathrm{2}}{\mathrm{3}βˆ’\mathrm{1}}=\mathrm{2} \\ $$$$\frac{\mathrm{6}}{{a}}=\mathrm{sin}\:\theta=\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}} \\ $$$$\Rightarrow{a}=\mathrm{3}\sqrt{\mathrm{5}} \\ $$$${area}\:{of}\:{square}\:={a}^{\mathrm{2}} =\left(\mathrm{3}\sqrt{\mathrm{5}}\right)^{\mathrm{2}} =\mathrm{45} \\ $$
Commented by necx122 last updated on 23/Jul/23
Thank you so much. Its very clear.
$${Thank}\:{you}\:{so}\:{much}.\:{Its}\:{very}\:{clear}. \\ $$

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