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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°

sin120°BC=sin20°AB⇒AB=23BCsin20°3sin20°AB=sin40°AC⇒AC=2ABcos20°=23BCsin403sin(x)DC=sinADCAC=sin(140−x)AC⇒ACDC=sin(140−x)sin(x)=sin40sin20=2cos20=sin140cosx−sin(x)cos(140)sinx=sin40tan(x)−cos(90+50)=sin40tanx+sin50⇒tanx=2cos20−sin502sin20cos20⇒x=60°

Commented by Rupesh123 last updated on 08/Nov/23

Perfect, sir!

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