Question Number 191856 by TUN last updated on 02/May/23
Answered by Subhi last updated on 02/May/23
![put AD=y ((BD)/(sin(30)))=(y/(sin(10))) BD=((sin(30))/(sin(10)))y by[sin law] ((CD)/(sin(50)))=(y/(sin(x))) .hence CD=((sin(50))/(sin(x)))y in triangleBDC ((BD)/(sin(70−x)))=((DC)/(sin(20))) ((sin(30).y)/(sin(10).sin(70−x)))=((sin(50).y)/(sin(20).sin(x))) (sin(30).sin(20))/(sin(50).sin(10)).sin(x)=sin(70).cos(x)−cos(70).sin(x) ((sin(30).sin(20))/(sin(50).sin(10).sin(70)))+cot(70)=cot(x) cot(x)=(√3) x=30](https://www.tinkutara.com/question/Q191864.png)
$$ \\ $$$${put}\:{AD}={y} \\ $$$$\frac{{BD}}{{sin}\left(\mathrm{30}\right)}=\frac{{y}}{{sin}\left(\mathrm{10}\right)} \\ $$$${BD}=\frac{{sin}\left(\mathrm{30}\right)}{{sin}\left(\mathrm{10}\right)}{y}\:\:\:\:\:\:\:{by}\left[{sin}\:{law}\right] \\ $$$$\frac{{CD}}{{sin}\left(\mathrm{50}\right)}=\frac{{y}}{{sin}\left({x}\right)}\:.{hence}\:{CD}=\frac{{sin}\left(\mathrm{50}\right)}{{sin}\left({x}\right)}{y} \\ $$$${in}\:{triangleBDC} \\ $$$$\frac{{BD}}{{sin}\left(\mathrm{70}−{x}\right)}=\frac{{DC}}{{sin}\left(\mathrm{20}\right)} \\ $$$$\frac{{sin}\left(\mathrm{30}\right).{y}}{{sin}\left(\mathrm{10}\right).{sin}\left(\mathrm{70}−{x}\right)}=\frac{{sin}\left(\mathrm{50}\right).{y}}{{sin}\left(\mathrm{20}\right).{sin}\left({x}\right)} \\ $$$$\left({sin}\left(\mathrm{30}\right).{sin}\left(\mathrm{20}\right)\right)/\left({sin}\left(\mathrm{50}\right).{sin}\left(\mathrm{10}\right)\right).{sin}\left({x}\right)={sin}\left(\mathrm{70}\right).{cos}\left({x}\right)−{cos}\left(\mathrm{70}\right).{sin}\left({x}\right) \\ $$$$\frac{{sin}\left(\mathrm{30}\right).{sin}\left(\mathrm{20}\right)}{{sin}\left(\mathrm{50}\right).{sin}\left(\mathrm{10}\right).{sin}\left(\mathrm{70}\right)}+{cot}\left(\mathrm{70}\right)={cot}\left({x}\right) \\ $$$${cot}\left({x}\right)=\sqrt{\mathrm{3}} \\ $$$${x}=\mathrm{30} \\ $$$$ \\ $$