Question Number 170012 by cortano1 last updated on 14/May/22
Answered by som(math1967) last updated on 14/May/22
![From △ABC ((AC)/(sin50))=((BC)/(sin30)) [∵∡ABC=180−100−30=50] from △DBC ((BD)/(sin100))=((BC)/(sinx)) ((BD)/(sin50))=((BC)/(sinxsin50))×sin(180−80) ((BD)/(sin50))=((sin80BC)/(sinxsin50)) ((AC)/(sin50))=((sin80BC)/(sinxsin50)) ((BC)/(sin30))=((2sin40cos40BC)/(sinxcos40)) sinx=sin40 ∴x=40](https://www.tinkutara.com/question/Q170013.png)
$${From}\:\bigtriangleup{ABC} \\ $$$$\:\frac{{AC}}{{sin}\mathrm{50}}=\frac{{BC}}{{sin}\mathrm{30}}\:\left[\because\measuredangle{ABC}=\mathrm{180}−\mathrm{100}−\mathrm{30}=\mathrm{50}\right] \\ $$$${from}\:\bigtriangleup{DBC} \\ $$$$\:\frac{{BD}}{{sin}\mathrm{100}}=\frac{{BC}}{{sinx}} \\ $$$$\:\frac{{BD}}{{sin}\mathrm{50}}=\frac{{BC}}{{sinxsin}\mathrm{50}}×{sin}\left(\mathrm{180}−\mathrm{80}\right) \\ $$$$\:\frac{{BD}}{{sin}\mathrm{50}}=\frac{{sin}\mathrm{80}{BC}}{{sinxsin}\mathrm{50}} \\ $$$$\:\frac{{AC}}{{sin}\mathrm{50}}=\frac{{sin}\mathrm{80}{BC}}{{sinxsin}\mathrm{50}} \\ $$$$\:\frac{{BC}}{{sin}\mathrm{30}}=\frac{\mathrm{2}{sin}\mathrm{40}{cos}\mathrm{40}{BC}}{{sinxcos}\mathrm{40}} \\ $$$${sinx}={sin}\mathrm{40}\:\therefore{x}=\mathrm{40} \\ $$