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Question Number 163292 by amin96 last updated on 05/Jan/22

Answered by mr W last updated on 05/Jan/22

say CD=AD=1 AC=2 cos 10° ((BC)/(sin 20))=((AC)/(sin (30+20))) ⇒BC=((2 cos 10 sin 20)/(sin 50)) ((sin x)/(BC))=((sin (180−x−20))/(CD)) ((sin 50)/(2 cos 10 sin 20))=((sin (x+20))/(sin x)) ((sin 50)/(2 cos 10 sin 20))=cos 20+((sin 20)/(tan x)) tan x=((sin 20)/(((sin 50)/(2 cos 10 sin 20))−cos 20))=(√3) ⇒x=60°

$${say}\:{CD}={AD}=\mathrm{1} \\ $$$${AC}=\mathrm{2}\:\mathrm{cos}\:\mathrm{10}° \\ $$$$\frac{{BC}}{\mathrm{sin}\:\mathrm{20}}=\frac{{AC}}{\mathrm{sin}\:\left(\mathrm{30}+\mathrm{20}\right)} \\ $$$$\Rightarrow{BC}=\frac{\mathrm{2}\:\mathrm{cos}\:\mathrm{10}\:\mathrm{sin}\:\mathrm{20}}{\mathrm{sin}\:\mathrm{50}} \\ $$$$\frac{\mathrm{sin}\:{x}}{{BC}}=\frac{\mathrm{sin}\:\left(\mathrm{180}−{x}−\mathrm{20}\right)}{{CD}} \\ $$$$\frac{\mathrm{sin}\:\mathrm{50}}{\mathrm{2}\:\mathrm{cos}\:\mathrm{10}\:\mathrm{sin}\:\mathrm{20}}=\frac{\mathrm{sin}\:\left({x}+\mathrm{20}\right)}{\mathrm{sin}\:{x}} \\ $$$$\frac{\mathrm{sin}\:\mathrm{50}}{\mathrm{2}\:\mathrm{cos}\:\mathrm{10}\:\mathrm{sin}\:\mathrm{20}}=\mathrm{cos}\:\mathrm{20}+\frac{\mathrm{sin}\:\mathrm{20}}{\mathrm{tan}\:{x}} \\ $$$$\mathrm{tan}\:{x}=\frac{\mathrm{sin}\:\mathrm{20}}{\frac{\mathrm{sin}\:\mathrm{50}}{\mathrm{2}\:\mathrm{cos}\:\mathrm{10}\:\mathrm{sin}\:\mathrm{20}}−\mathrm{cos}\:\mathrm{20}}=\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{x}=\mathrm{60}° \\ $$

Commented by Tawa11 last updated on 06/Jan/22

Great sir

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

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