Question Number 195742 by sonukgindia last updated on 09/Aug/23
Answered by MM42 last updated on 09/Aug/23
$$\angle\alpha_{{i}} =\mathrm{120}\Rightarrow\angle{BAC}=\angle{BDF}=\mathrm{90} \\ $$$${AB}={AC}\Rightarrow\angle{C}=\angle{B}_{\mathrm{1},\mathrm{2}} =\mathrm{45}\Rightarrow\angle{B}_{\mathrm{2}} =\angle{B}_{\mathrm{3}} =\mathrm{15} \\ $$$$\frac{{x}}{{sin}\mathrm{15}}=\frac{{BE}}{{sin}\mathrm{90}}\:\:\:\&\:\:\:\frac{{y}}{{sin}\mathrm{15}}=\frac{{BE}}{{sin}\mathrm{60}} \\ $$$$\Rightarrow\frac{{x}}{{y}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:\checkmark \\ $$$$ \\ $$