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Question Number 177980 by infinityaction last updated on 11/Oct/22

Answered by mr W last updated on 11/Oct/22

Commented by mr W last updated on 11/Oct/22

((EB)/(sin 144°))=(1/(sin x)) ((EA)/(sin 36°))=(1/(sin 42°)) ((EB)/(sin 84°))=((EA)/(sin (30°−x))) ((sin 144°)/(sin x sin 84°))=((sin 36°)/(sin 42° sin (30°−x))) ((sin (30°−x))/(sin x))=((sin 84°)/(sin 42°))=2 cos 42° (1/(2 tan x))−((√3)/2)=2 cos 42° tan x=(1/(4 cos 42°+(√3))) ⇒x=12°

$$\frac{{EB}}{\mathrm{sin}\:\mathrm{144}°}=\frac{\mathrm{1}}{\mathrm{sin}\:{x}} \\ $$$$\frac{{EA}}{\mathrm{sin}\:\mathrm{36}°}=\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{42}°} \\ $$$$\frac{{EB}}{\mathrm{sin}\:\mathrm{84}°}=\frac{{EA}}{\mathrm{sin}\:\left(\mathrm{30}°−{x}\right)} \\ $$$$\frac{\mathrm{sin}\:\mathrm{144}°}{\mathrm{sin}\:{x}\:\mathrm{sin}\:\mathrm{84}°}=\frac{\mathrm{sin}\:\mathrm{36}°}{\mathrm{sin}\:\mathrm{42}°\:\mathrm{sin}\:\left(\mathrm{30}°−{x}\right)} \\ $$$$\frac{\mathrm{sin}\:\left(\mathrm{30}°−{x}\right)}{\mathrm{sin}\:{x}}=\frac{\mathrm{sin}\:\mathrm{84}°}{\mathrm{sin}\:\mathrm{42}°}=\mathrm{2}\:\mathrm{cos}\:\mathrm{42}° \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{tan}\:{x}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{2}\:\mathrm{cos}\:\mathrm{42}° \\ $$$$\mathrm{tan}\:{x}=\frac{\mathrm{1}}{\mathrm{4}\:\mathrm{cos}\:\mathrm{42}°+\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{x}=\mathrm{12}° \\ $$

Commented by Tawa11 last updated on 11/Oct/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Commented by infinityaction last updated on 12/Oct/22

thanks sir

$${thanks}\:{sir} \\ $$

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