Question Number 193958 by Rupesh123 last updated on 24/Jun/23
Answered by ARUNG_Brandon_MBU last updated on 24/Jun/23
$$\frac{{a}}{\mathrm{sin}\left(\mathrm{150}°−{x}\right)}=\frac{{b}}{\mathrm{sin30}°}=\mathrm{2}{b} \\ $$$$\frac{{a}}{\mathrm{sin30}°}=\frac{{a}+{b}}{\mathrm{sin}\left(\mathrm{150}°−{x}\right)}=\mathrm{2}{a} \\ $$$$\Rightarrow\mathrm{2}{b}\left({a}+{b}\right)=\mathrm{2}{a}^{\mathrm{2}} \:\Rightarrow{a}^{\mathrm{2}} −{ab}−{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\left(\frac{{a}}{{b}}\right)^{\mathrm{2}} −\frac{{a}}{{b}}−\mathrm{1}=\mathrm{0}\:\Rightarrow\frac{{a}}{{b}}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\left(\mathrm{150}°−{x}\right)=\frac{{a}}{\mathrm{2}{b}}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{sin}\left({x}+\mathrm{30}°\right)=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$\Rightarrow{x}=\mathrm{arcsin}\left(\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{4}}\right)−\mathrm{30}°\:,\:{x}>\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{arcsin}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\right)−\mathrm{30}°=\mathrm{24}° \\ $$
Commented by Rupesh123 last updated on 25/Jun/23
Perfect