Question Number 194029 by Rupesh123 last updated on 26/Jun/23
Commented by Rupesh123 last updated on 26/Jun/23
Perfect ��
Answered by Subhi last updated on 26/Jun/23
$${AD}\:=\:\sqrt{\mathrm{2}{Q}^{\mathrm{2}} −\mathrm{2}{Q}^{\mathrm{2}} {cos}\left(\mathrm{120}\right)}=\sqrt{\mathrm{3}}{Q} \\ $$$${AB}\:=\:{AC}\:=\:{Z} \\ $$$$\frac{{Z}}{{sin}\left(\mathrm{150}\right)}=\frac{\sqrt{\mathrm{3}}{Q}}{{sin}\left(\mathrm{20}\right)}\:\Rrightarrow\:{Q}\:=\:\frac{\mathrm{2}{sin}\left(\mathrm{20}\right){Z}}{\:\sqrt{\mathrm{3}}} \\ $$$$\frac{\mathrm{2}{sin}\left(\mathrm{20}\right){Z}}{\:\sqrt{\mathrm{3}}.{sin}\left(\mathrm{120}−{x}\right)}=\frac{{z}−\frac{\mathrm{2}{sin}\left(\mathrm{120}\right){Z}}{\:\sqrt{\mathrm{3}}}}{{sin}\left({x}\right)} \\ $$$$\mathrm{2}{sin}\left(\mathrm{20}\right){sin}\left({x}\right)=\left(\frac{\sqrt{\mathrm{3}}{cos}\left({x}\right)}{\mathrm{2}}+\frac{{sin}\left({x}\right)}{\mathrm{2}}\right)\left(\sqrt{\mathrm{3}}−\mathrm{2}{sin}\left(\mathrm{20}\right)\right) \\ $$$${x}\:=\:{tan}^{−\mathrm{1}} \left(\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}−\mathrm{2}{sin}\left(\mathrm{20}\right)\right)}{\mathrm{2}{sin}\left(\mathrm{20}\right)−\frac{\sqrt{\mathrm{3}}−\mathrm{2}{sin}\left(\mathrm{20}\right)}{\mathrm{2}}}\right)=\mathrm{80} \\ $$
Commented by Subhi last updated on 26/Jun/23
