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
$$\mathrm{sin}\:\theta=\frac{\mathrm{21}}{\mathrm{49}}=\frac{\mathrm{3}}{\mathrm{7}} \\ $$$$\mathrm{cos}\:\mathrm{2}\theta=\mathrm{1}−\mathrm{2}×\left(\frac{\mathrm{3}}{\mathrm{7}}\right)^{\mathrm{2}} =\frac{\mathrm{31}}{\mathrm{49}} \\ $$$$\left(\mathrm{2}{x}\right)^{\mathrm{2}} =\mathrm{49}^{\mathrm{2}} +\mathrm{85}^{\mathrm{2}} −\mathrm{2}×\mathrm{49}×\mathrm{85}×\frac{\mathrm{31}}{\mathrm{49}}=\mathrm{4356} \\ $$$$\Rightarrow{x}=\frac{\sqrt{\mathrm{4356}}}{\mathrm{2}}=\mathrm{33}\:\checkmark \\ $$
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
$$\mathrm{cos}\:\theta=\frac{\mathrm{21}}{\mathrm{49}} \\ $$$${x}^{\mathrm{2}} =\mathrm{21}^{\mathrm{2}} +\mathrm{18}^{\mathrm{2}} +\mathrm{2}×\mathrm{21}×\mathrm{18}×\frac{\mathrm{21}}{\mathrm{49}}=\mathrm{1089} \\ $$$$\Rightarrow{x}=\sqrt{\mathrm{1089}}=\mathrm{33} \\ $$
Commented by Rupesh123 last updated on 15/Feb/23
Good job!
Answered by a.lgnaoui last updated on 14/Feb/23
$${there}\:{more}\:{methodes} \\ $$$${Simple}\:{methode} \\ $$$$\bigtriangleup{ABC}\:{ANE}\left({Semblables}\right){BC}//{DE} \\ $$$$\:\frac{{AB}}{{AN}}=\frac{{AC}}{{AE}}=\frac{{BC}}{{NE}}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{NE}=\mathrm{36} \\ $$$$\:\:\Rightarrow{BN}=\mathrm{21}\:\:\:\:{AN}=\mathrm{42} \\ $$$$\bigtriangleup{ANE}\:\: \\ $$$${AN}^{\mathrm{2}} ={AE}^{\mathrm{2}} +{NE}^{\mathrm{2}} −\mathrm{2}{AE}×{NE}\mathrm{cos}\:\lambda\left(\mathrm{1}\right) \\ $$$$\mathrm{42}^{\mathrm{2}} =\mathrm{36}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}×\mathrm{36cos}\:\lambda \\ $$$$=\mathrm{144}{x}\mathrm{cos}\:\lambda=\mathrm{4}{x}^{\mathrm{2}} −\mathrm{468} \\ $$$$\mathrm{cos}\:\lambda=\frac{{x}^{\mathrm{2}} −\mathrm{117}}{\mathrm{36}{x}}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\bigtriangleup{ADN}\:\:{AB}\:{et}\:{perpendiculaires} \\ $$$${BD}^{\mathrm{2}} +{BN}^{\mathrm{2}} ={DN}^{\mathrm{2}} \\ $$$${DE}=\mathrm{85}={DN}+{NE}={DN}+\mathrm{36}\Rightarrow{DN}=\mathrm{49}\left(\mathrm{2}\right) \\ $$$${DN}=\mathrm{49} \\ $$$$\bigtriangleup{ADE} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\Rightarrow{AD}^{\mathrm{2}} ={DE}^{\mathrm{2}} +{AE}^{\mathrm{2}} −{DE}×{AE}\mathrm{cos}\:\lambda \\ $$$$\mathrm{49}^{\mathrm{2}} =\mathrm{85}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}×\mathrm{85}{x}\mathrm{cos}\:\lambda \\ $$$$\mathrm{49}^{\mathrm{2}} =\mathrm{85}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{340}{x}\mathrm{cos}\:\lambda \\ $$$$\:\mathrm{340}{x}\mathrm{cos}\:\lambda=\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4824} \\ $$$$\mathrm{cos}\:\lambda=\frac{{x}^{\mathrm{2}} +\mathrm{1206}}{\mathrm{85}{x}}\:\:\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$${AB}={BN}\Rightarrow\bigtriangleup{ADN}:\:{Isocele}\:\left({AD}={DN}\right) \\ $$$$ \\ $$$$\bigtriangleup{ADE}\:\:\: \\ $$$${AD}^{\mathrm{2}} ={AE}^{\mathrm{2}} +{DE}^{\mathrm{2}} −\mathrm{2}{AE}×{DE}\mathrm{cos}\:\lambda \\ $$$$\:\:\:\:\mathrm{49}^{\mathrm{2}} =\mathrm{4}{x}^{\mathrm{2}} +\mathrm{85}^{\mathrm{2}} −\mathrm{340}{x}\mathrm{cos}\:\lambda\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\:\:\left(\mathrm{1}\right)\:{et}\:\left(\mathrm{2}\right)\Rightarrow\mathrm{cos}\:\lambda=\frac{{x}^{\mathrm{2}} +\mathrm{1206}}{\mathrm{85}{x}}=\frac{{x}^{\mathrm{2}} −\mathrm{117}}{\mathrm{36}{x}} \\ $$$$\:\frac{{x}^{\mathrm{2}} +\mathrm{1206}}{\mathrm{85}}=\frac{{x}^{\mathrm{2}} −\mathrm{117}}{\mathrm{36}} \\ $$$$\mathrm{85}\left({x}^{\mathrm{2}} −\mathrm{117}\right)=\mathrm{36}\left({x}^{\mathrm{2}} +\mathrm{1206}\right) \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{85}×\mathrm{117}+\mathrm{36}×\mathrm{1206}}{\mathrm{85}} \\ $$$${Resultat}\:{definitif}\:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} =\frac{\mathrm{9945}+\mathrm{43416}}{\mathrm{49}}=\mathrm{1089} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{x}=\mathrm{33} \\ $$$$\: \\ $$
Commented by a.lgnaoui last updated on 14/Feb/23
Commented by mr W last updated on 14/Feb/23
$${yes},\:{there}\:{are}\:{many}\:{right}\:{methods}\:{and} \\ $$$${there}\:{are}\:{also}\:{many}\:{wrong}\:{methods}. \\ $$
Commented by mr W last updated on 14/Feb/23
$${now}\:{your}\:{answer}\:{is}\:{correct}.\:{when}\:{you} \\ $$$${think}\:{your}\:{solution}\:{is}\:{simple},\:{it}'{s} \\ $$$${your}\:{right}\:{to}\:{think}\:{what}\:{you}\:{think}. \\ $$
Commented by a.lgnaoui last updated on 14/Feb/23
$${rectification}\:{calcul} \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{43416}+\mathrm{9945}}{\mathrm{49}}=\frac{\mathrm{53361}}{\mathrm{49}} \\ $$$${x}=\mathrm{33} \\ $$
Commented by a.lgnaoui last updated on 14/Feb/23
$${its}\:{only}\:\:{probleme}\:{operation} \\ $$$${calcul} \\ $$
Commented by a.lgnaoui last updated on 14/Feb/23
$${its}\:{a}\:{simple}\:{probleme}\:{calcul} \\ $$$${sorry}. \\ $$
Commented by ajfour last updated on 14/Feb/23
$${dont}\:{u}\:{see}\:{the}\:{font}\:{size}\:{u}\:{use}, \\ $$$${how}\:{do}\:{you}\:{type}\:{so}\:{big}? \\ $$$${guess}\:{i}\:{knew}..! \\ $$
Commented by a.lgnaoui last updated on 14/Feb/23
$${yes}\:{ok} \\ $$
Commented by Rupesh123 last updated on 15/Feb/23
Good job!