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Question Number 199728 by Mingma last updated on 08/Nov/23

Answered by deleteduser1 last updated on 08/Nov/23

((sin120°)/(BC))=((sin20°)/(AB))⇒AB=((2(√3)BCsin20°)/3)  ((sin20°)/(AB))=((sin40°)/(AC))⇒AC=2ABcos20°=((2(√3)BCsin40)/3)  ((sin(x))/(DC))=((sinADC)/(AC))=((sin(140−x))/(AC))⇒((AC)/(DC))=((sin(140−x))/(sin(x)))  =((sin40)/(sin20))=2cos20=((sin140cosx−sin(x)cos(140))/(sinx))  =sin40tan(x)−cos(90+50)=sin40tanx+sin50  ⇒tanx=((2cos20−sin50)/(2sin20cos20))⇒x=60°

$$\frac{{sin}\mathrm{120}°}{{BC}}=\frac{{sin}\mathrm{20}°}{{AB}}\Rightarrow{AB}=\frac{\mathrm{2}\sqrt{\mathrm{3}}{BCsin}\mathrm{20}°}{\mathrm{3}} \\ $$$$\frac{{sin}\mathrm{20}°}{{AB}}=\frac{{sin}\mathrm{40}°}{{AC}}\Rightarrow{AC}=\mathrm{2}{ABcos}\mathrm{20}°=\frac{\mathrm{2}\sqrt{\mathrm{3}}{BCsin}\mathrm{40}}{\mathrm{3}} \\ $$$$\frac{{sin}\left({x}\right)}{{DC}}=\frac{{sinADC}}{{AC}}=\frac{{sin}\left(\mathrm{140}−{x}\right)}{{AC}}\Rightarrow\frac{{AC}}{{DC}}=\frac{{sin}\left(\mathrm{140}−{x}\right)}{{sin}\left({x}\right)} \\ $$$$=\frac{{sin}\mathrm{40}}{{sin}\mathrm{20}}=\mathrm{2}{cos}\mathrm{20}=\frac{{sin}\mathrm{140}{cosx}−{sin}\left({x}\right){cos}\left(\mathrm{140}\right)}{{sinx}} \\ $$$$={sin}\mathrm{40}{tan}\left({x}\right)−{cos}\left(\mathrm{90}+\mathrm{50}\right)={sin}\mathrm{40}{tanx}+{sin}\mathrm{50} \\ $$$$\Rightarrow{tanx}=\frac{\mathrm{2}{cos}\mathrm{20}−{sin}\mathrm{50}}{\mathrm{2}{sin}\mathrm{20}{cos}\mathrm{20}}\Rightarrow{x}=\mathrm{60}° \\ $$

Commented by Rupesh123 last updated on 08/Nov/23

Perfect, sir!

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