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Question Number 196275 by AROUNAMoussa last updated on 21/Aug/23

Answered by MM42 last updated on 21/Aug/23

⟨ABD=35⇒AD=AB ⟨BCD=55 ((AB)/(AC))=((sinx)/(sin130)) & ((AD)/(AC))=((sin(55−x))/(sin65)) ⇒sinx×sin65=2sin65×cos65×sin(55−x) ⇒sinx=2cos65sin(55−x)=sin(55−x+65)+sin(55−x−65) ⇒sinx=sin(120−x)−sin(x+10) ⇒sinx=((√3)/2)cosx−(1/2)sinx−cos10sinx−sin10cosx ((3/2)−cos10)sinx=(((√3)/2)−sin10)cosx tanx=(((√3)−2sin10)/(3−2cos10)) ⇒x=tan^(−1) ((((√3)−2sin10)/(3−2cos10)))∼53^o

$$\langle{ABD}=\mathrm{35}\Rightarrow{AD}={AB} \\ $$$$\langle{BCD}=\mathrm{55} \\ $$$$\frac{{AB}}{{AC}}=\frac{{sinx}}{{sin}\mathrm{130}}\:\:\:\&\:\:\frac{{AD}}{{AC}}=\frac{{sin}\left(\mathrm{55}−{x}\right)}{{sin}\mathrm{65}} \\ $$$$\Rightarrow{sinx}×{sin}\mathrm{65}=\mathrm{2}{sin}\mathrm{65}×{cos}\mathrm{65}×{sin}\left(\mathrm{55}−{x}\right) \\ $$$$\Rightarrow{sinx}=\mathrm{2}{cos}\mathrm{65}{sin}\left(\mathrm{55}−{x}\right)={sin}\left(\mathrm{55}−{x}+\mathrm{65}\right)+{sin}\left(\mathrm{55}−{x}−\mathrm{65}\right) \\ $$$$\Rightarrow{sinx}={sin}\left(\mathrm{120}−{x}\right)−{sin}\left({x}+\mathrm{10}\right) \\ $$$$\Rightarrow{sinx}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{cosx}−\frac{\mathrm{1}}{\mathrm{2}}{sinx}−{cos}\mathrm{10}{sinx}−{sin}\mathrm{10}{cosx} \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{2}}−{cos}\mathrm{10}\right){sinx}=\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−{sin}\mathrm{10}\right){cosx} \\ $$$${tanx}=\frac{\sqrt{\mathrm{3}}−\mathrm{2}{sin}\mathrm{10}}{\mathrm{3}−\mathrm{2}{cos}\mathrm{10}} \\ $$$$\:\Rightarrow{x}={tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{3}}−\mathrm{2}{sin}\mathrm{10}}{\mathrm{3}−\mathrm{2}{cos}\mathrm{10}}\right)\sim\mathrm{53}^{{o}} \\ $$$$ \\ $$$$ \\ $$

Commented by AROUNAMoussa last updated on 21/Aug/23

Merci bcp !

$${Merci}\:{bcp}\:! \\ $$

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