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Question Number 185770 by Mingma last updated on 27/Jan/23

Answered by mahdipoor last updated on 27/Jan/23

$$\mathrm{\Delta }ADB:\frac{AB}{sin(180-(20+x))}=\frac{BD}{sin\left(x\right)}$$$$\mathrm{\Delta }BDC:\frac{BD}{sin30}=\frac{BC}{sin(180-(50+30))}$$$$\mathrm{\Delta }ABC:\frac{AB}{sin70}=\frac{BC}{sin(180-(70+70))}$$$$\Rightarrow \Rightarrow BD=\frac{sin\left(x\right)}{sin(20+x)}AB=\frac{sin\left(30\right)}{sin\left(80\right)}BC=$$$$\frac{sin\left(30\right)}{sin\left(80\right)}\times \frac{sin\left(40\right)}{sin\left(70\right)}AB$$$$\Rightarrow \frac{sin\left(x\right)}{sin(x+20)}=\frac{sin\left(40\right).sin\left(30\right)}{sin\left(70\right).sin\left(80\right)}\Rightarrow$$$$ctg\left(x\right)=\frac{sin\left(70\right)sin\left(80\right)}{sin\left(40\right)sin\left(30\right)sin\left(20\right)}-ctg\left(20\right)$$$$\Rightarrow x=10$$

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