Question Number 187100 by Tons last updated on 13/Feb/23
Answered by a.lgnaoui last updated on 14/Feb/23
$$\bigtriangleup{APB}\:\:\:\:{AB}\mathrm{sin}\:{X}={AC}\mathrm{cos}\:{Y}\:\: \\ $$$$ \\ $$$$\:{BC}=\mathrm{2}{AB}\mathrm{cos}\:{Y}\:\:\Rightarrow\begin{cases}{{Y}=\frac{\pi}{\mathrm{2}}−{X}}\\{{AB}={AC}}\end{cases} \\ $$$$\mathrm{sin}\:{X}=\mathrm{cos}\:{Y}\:\:\: \\ $$$$\bigtriangleup{ABCD}\:\:\:\:{Sqart}\left({Care}\right) \\ $$$$\:\:\:\:\mathrm{2}{Y}=\frac{\pi}{\mathrm{2}}\Rightarrow\:\:\:\:\:{Y}=\frac{\pi}{\mathrm{4}}\:\:\:\:;{X}=\frac{\pi}{\mathrm{4}} \\ $$$$ \\ $$
Commented by a.lgnaoui last updated on 14/Feb/23
