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

Answered by mahdipoor last updated on 27/Jan/23

ΔADB: ((AB)/(sin(180−(20+x))))=((BD)/(sin(x))) ΔBDC: ((BD)/(sin30))=((BC)/(sin(180−(50+30)))) ΔABC: ((AB)/(sin70))=((BC)/(sin(180−(70+70)))) ⇒⇒BD=((sin(x))/(sin(20+x)))AB=((sin(30))/(sin(80)))BC= ((sin(30))/(sin(80)))×((sin(40))/(sin(70)))AB ⇒((sin(x))/(sin(x+20)))=((sin(40).sin(30))/(sin(70).sin(80)))⇒ ctg(x)=((sin(70)sin(80))/(sin(40)sin(30)sin(20)))−ctg(20) ⇒x=10

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

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