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Question Number 187167 by Rupesh123 last updated on 14/Feb/23

Answered by mr W last updated on 14/Feb/23

Commented by mr W last updated on 14/Feb/23

sin θ=((21)/(49))=(3/7) cos 2θ=1−2×((3/7))^2 =((31)/(49)) (2x)^2 =49^2 +85^2 −2×49×85×((31)/(49))=4356 ⇒x=((√(4356))/2)=33 ✓

$$\mathrm{sin}\:\theta=\frac{\mathrm{21}}{\mathrm{49}}=\frac{\mathrm{3}}{\mathrm{7}} \\ $$$$\mathrm{cos}\:\mathrm{2}\theta=\mathrm{1}−\mathrm{2}×\left(\frac{\mathrm{3}}{\mathrm{7}}\right)^{\mathrm{2}} =\frac{\mathrm{31}}{\mathrm{49}} \\ $$$$\left(\mathrm{2}{x}\right)^{\mathrm{2}} =\mathrm{49}^{\mathrm{2}} +\mathrm{85}^{\mathrm{2}} −\mathrm{2}×\mathrm{49}×\mathrm{85}×\frac{\mathrm{31}}{\mathrm{49}}=\mathrm{4356} \\ $$$$\Rightarrow{x}=\frac{\sqrt{\mathrm{4356}}}{\mathrm{2}}=\mathrm{33}\:\checkmark \\ $$

Commented by Rupesh123 last updated on 14/Feb/23

Nice!

Answered by mr W last updated on 14/Feb/23

Commented by mr W last updated on 14/Feb/23

cos θ=((21)/(49)) x^2 =21^2 +18^2 +2×21×18×((21)/(49))=1089 ⇒x=(√(1089))=33

$$\mathrm{cos}\:\theta=\frac{\mathrm{21}}{\mathrm{49}} \\ $$$${x}^{\mathrm{2}} =\mathrm{21}^{\mathrm{2}} +\mathrm{18}^{\mathrm{2}} +\mathrm{2}×\mathrm{21}×\mathrm{18}×\frac{\mathrm{21}}{\mathrm{49}}=\mathrm{1089} \\ $$$$\Rightarrow{x}=\sqrt{\mathrm{1089}}=\mathrm{33} \\ $$

Commented by Rupesh123 last updated on 15/Feb/23

Good job!

Answered by a.lgnaoui last updated on 14/Feb/23

there more methodes Simple methode △ABC ANE(Semblables)BC//DE ((AB)/(AN))=((AC)/(AE))=((BC)/(NE))=(1/2)⇒NE=36 ⇒BN=21 AN=42 △ANE AN^2 =AE^2 +NE^2 −2AE×NEcos λ(1) 42^2 =36^2 +4x^2 −4x×36cos λ =144xcos λ=4x^2 −468 cos λ=((x^2 −117)/(36x)) (1) △ADN AB et perpendiculaires BD^2 +BN^2 =DN^2 DE=85=DN+NE=DN+36⇒DN=49(2) DN=49 △ADE (1)⇒AD^2 =DE^2 +AE^2 −DE×AEcos λ 49^2 =85^2 +4x^2 −4×85xcos λ 49^2 =85^2 +4x^2 −340xcos λ 340xcos λ=4x^2 +4824 cos λ=((x^2 +1206)/(85x)) (2) AB=BN⇒△ADN: Isocele (AD=DN) △ADE AD^2 =AE^2 +DE^2 −2AE×DEcos λ 49^2 =4x^2 +85^2 −340xcos λ (2) (1) et (2)⇒cos λ=((x^2 +1206)/(85x))=((x^2 −117)/(36x)) ((x^2 +1206)/(85))=((x^2 −117)/(36)) 85(x^2 −117)=36(x^2 +1206) x^2 =((85×117+36×1206)/(85)) Resultat definitif : x^2 =((9945+43416)/(49))=1089 x=33

$${there}\:{more}\:{methodes} \\ $$$${Simple}\:{methode} \\ $$$$\bigtriangleup{ABC}\:{ANE}\left({Semblables}\right){BC}//{DE} \\ $$$$\:\frac{{AB}}{{AN}}=\frac{{AC}}{{AE}}=\frac{{BC}}{{NE}}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{NE}=\mathrm{36} \\ $$$$\:\:\Rightarrow{BN}=\mathrm{21}\:\:\:\:{AN}=\mathrm{42} \\ $$$$\bigtriangleup{ANE}\:\: \\ $$$${AN}^{\mathrm{2}} ={AE}^{\mathrm{2}} +{NE}^{\mathrm{2}} −\mathrm{2}{AE}×{NE}\mathrm{cos}\:\lambda\left(\mathrm{1}\right) \\ $$$$\mathrm{42}^{\mathrm{2}} =\mathrm{36}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}×\mathrm{36cos}\:\lambda \\ $$$$=\mathrm{144}{x}\mathrm{cos}\:\lambda=\mathrm{4}{x}^{\mathrm{2}} −\mathrm{468} \\ $$$$\mathrm{cos}\:\lambda=\frac{{x}^{\mathrm{2}} −\mathrm{117}}{\mathrm{36}{x}}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\bigtriangleup{ADN}\:\:{AB}\:{et}\:{perpendiculaires} \\ $$$${BD}^{\mathrm{2}} +{BN}^{\mathrm{2}} ={DN}^{\mathrm{2}} \\ $$$${DE}=\mathrm{85}={DN}+{NE}={DN}+\mathrm{36}\Rightarrow{DN}=\mathrm{49}\left(\mathrm{2}\right) \\ $$$${DN}=\mathrm{49} \\ $$$$\bigtriangleup{ADE} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\Rightarrow{AD}^{\mathrm{2}} ={DE}^{\mathrm{2}} +{AE}^{\mathrm{2}} −{DE}×{AE}\mathrm{cos}\:\lambda \\ $$$$\mathrm{49}^{\mathrm{2}} =\mathrm{85}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}×\mathrm{85}{x}\mathrm{cos}\:\lambda \\ $$$$\mathrm{49}^{\mathrm{2}} =\mathrm{85}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{340}{x}\mathrm{cos}\:\lambda \\ $$$$\:\mathrm{340}{x}\mathrm{cos}\:\lambda=\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4824} \\ $$$$\mathrm{cos}\:\lambda=\frac{{x}^{\mathrm{2}} +\mathrm{1206}}{\mathrm{85}{x}}\:\:\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$${AB}={BN}\Rightarrow\bigtriangleup{ADN}:\:{Isocele}\:\left({AD}={DN}\right) \\ $$$$ \\ $$$$\bigtriangleup{ADE}\:\:\: \\ $$$${AD}^{\mathrm{2}} ={AE}^{\mathrm{2}} +{DE}^{\mathrm{2}} −\mathrm{2}{AE}×{DE}\mathrm{cos}\:\lambda \\ $$$$\:\:\:\:\mathrm{49}^{\mathrm{2}} =\mathrm{4}{x}^{\mathrm{2}} +\mathrm{85}^{\mathrm{2}} −\mathrm{340}{x}\mathrm{cos}\:\lambda\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\:\:\left(\mathrm{1}\right)\:{et}\:\left(\mathrm{2}\right)\Rightarrow\mathrm{cos}\:\lambda=\frac{{x}^{\mathrm{2}} +\mathrm{1206}}{\mathrm{85}{x}}=\frac{{x}^{\mathrm{2}} −\mathrm{117}}{\mathrm{36}{x}} \\ $$$$\:\frac{{x}^{\mathrm{2}} +\mathrm{1206}}{\mathrm{85}}=\frac{{x}^{\mathrm{2}} −\mathrm{117}}{\mathrm{36}} \\ $$$$\mathrm{85}\left({x}^{\mathrm{2}} −\mathrm{117}\right)=\mathrm{36}\left({x}^{\mathrm{2}} +\mathrm{1206}\right) \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{85}×\mathrm{117}+\mathrm{36}×\mathrm{1206}}{\mathrm{85}} \\ $$$${Resultat}\:{definitif}\:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} =\frac{\mathrm{9945}+\mathrm{43416}}{\mathrm{49}}=\mathrm{1089} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{x}=\mathrm{33} \\ $$$$\: \\ $$

Commented by a.lgnaoui last updated on 14/Feb/23

Commented by mr W last updated on 14/Feb/23

yes, there are many right methods and there are also many wrong methods.

$${yes},\:{there}\:{are}\:{many}\:{right}\:{methods}\:{and} \\ $$$${there}\:{are}\:{also}\:{many}\:{wrong}\:{methods}. \\ $$

Commented by mr W last updated on 14/Feb/23

now your answer is correct. when you think your solution is simple, it′s your right to think what you think.

$${now}\:{your}\:{answer}\:{is}\:{correct}.\:{when}\:{you} \\ $$$${think}\:{your}\:{solution}\:{is}\:{simple},\:{it}'{s} \\ $$$${your}\:{right}\:{to}\:{think}\:{what}\:{you}\:{think}. \\ $$

Commented by a.lgnaoui last updated on 14/Feb/23

rectification calcul x^2 =((43416+9945)/(49))=((53361)/(49)) x=33

$${rectification}\:{calcul} \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{43416}+\mathrm{9945}}{\mathrm{49}}=\frac{\mathrm{53361}}{\mathrm{49}} \\ $$$${x}=\mathrm{33} \\ $$

Commented by a.lgnaoui last updated on 14/Feb/23

its only probleme operation calcul

$${its}\:{only}\:\:{probleme}\:{operation} \\ $$$${calcul} \\ $$

Commented by a.lgnaoui last updated on 14/Feb/23

its a simple probleme calcul sorry.

$${its}\:{a}\:{simple}\:{probleme}\:{calcul} \\ $$$${sorry}. \\ $$

Commented by ajfour last updated on 14/Feb/23

dont u see the font size u use, how do you type so big? guess i knew..!

$${dont}\:{u}\:{see}\:{the}\:{font}\:{size}\:{u}\:{use}, \\ $$$${how}\:{do}\:{you}\:{type}\:{so}\:{big}? \\ $$$${guess}\:{i}\:{knew}..! \\ $$

Commented by a.lgnaoui last updated on 14/Feb/23

yes ok

$${yes}\:{ok} \\ $$

Commented by Rupesh123 last updated on 15/Feb/23

Good job!

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