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Question Number 184370 by cherokeesay last updated on 05/Jan/23

Commented by cherokeesay last updated on 05/Jan/23

on demande la mesure de alpha :

Answered by som(math1967) last updated on 05/Jan/23

((AD)/(DC))=((sinα)/(sin50))  ((AD)/(AB))=((sin40)/(sin(50+α)))  ⇒((sinα)/(sin50))=((sin40)/(sin(50+α)))  ⇒2sinαsin(50+α)=2sin40sin50  ⇒cos50−cos(50+2α)=cos10−cos90  ⇒cos(50+α)=cos50−cos10  ⇒cos(50+α)=−sin20  ⇒cos(50+2α)=−cos70  ⇒cos(50+2α)=cos(180−70)  50+2α=110  α=30

$$\frac{{AD}}{{DC}}=\frac{{sin}\alpha}{{sin}\mathrm{50}} \\ $$$$\frac{{AD}}{{AB}}=\frac{{sin}\mathrm{40}}{{sin}\left(\mathrm{50}+\alpha\right)} \\ $$$$\Rightarrow\frac{{sin}\alpha}{{sin}\mathrm{50}}=\frac{{sin}\mathrm{40}}{{sin}\left(\mathrm{50}+\alpha\right)} \\ $$$$\Rightarrow\mathrm{2}{sin}\alpha{sin}\left(\mathrm{50}+\alpha\right)=\mathrm{2}{sin}\mathrm{40}{sin}\mathrm{50} \\ $$$$\Rightarrow{cos}\mathrm{50}−{cos}\left(\mathrm{50}+\mathrm{2}\alpha\right)={cos}\mathrm{10}−{cos}\mathrm{90} \\ $$$$\Rightarrow{cos}\left(\mathrm{50}+\alpha\right)={cos}\mathrm{50}−{cos}\mathrm{10} \\ $$$$\Rightarrow{cos}\left(\mathrm{50}+\alpha\right)=−{sin}\mathrm{20} \\ $$$$\Rightarrow{cos}\left(\mathrm{50}+\mathrm{2}\alpha\right)=−{cos}\mathrm{70} \\ $$$$\Rightarrow{cos}\left(\mathrm{50}+\mathrm{2}\alpha\right)={cos}\left(\mathrm{180}−\mathrm{70}\right) \\ $$$$\mathrm{50}+\mathrm{2}\alpha=\mathrm{110} \\ $$$$\alpha=\mathrm{30} \\ $$

Commented by cherokeesay last updated on 05/Jan/23

thank you sir !

$${thank}\:{you}\:{sir}\:! \\ $$

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