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Question Number 206198 by lmcp1203 last updated on 09/Apr/24

Answered by A5T last updated on 09/Apr/24

((sin4θ)/(AD))=((sin(90−3θ))/(AB));((sin(θ))/(AD))=((sin(90−4θ))/(CD=AB)) ⇒((sin(4θ))/(sin(90−3θ)))=((sin(θ))/(sin(90−4θ)))⇒x=20°

$$\frac{{sin}\mathrm{4}\theta}{{AD}}=\frac{{sin}\left(\mathrm{90}−\mathrm{3}\theta\right)}{{AB}};\frac{{sin}\left(\theta\right)}{{AD}}=\frac{{sin}\left(\mathrm{90}−\mathrm{4}\theta\right)}{{CD}={AB}} \\ $$$$\Rightarrow\frac{{sin}\left(\mathrm{4}\theta\right)}{{sin}\left(\mathrm{90}−\mathrm{3}\theta\right)}=\frac{{sin}\left(\theta\right)}{{sin}\left(\mathrm{90}−\mathrm{4}\theta\right)}\Rightarrow{x}=\mathrm{20}° \\ $$

Answered by lmcp1203 last updated on 09/Apr/24

thanks.

$${thanks}. \\ $$

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