Hei! :)
- ΔABC isoscel
- AB=AC= 18 cm
- ∡ B=72
a) BC=?
Fie AE mijlocul lui [BC] si BD mijlocul [AC]
=> ΔBCD asemenea cu ΔABC
=> [tex]\frac{AB}{BC} =\frac{BC}{CD} <=>\frac{18}{x} =\frac{x}{18-x} \\=> x^{2} =18*(18-x)\\x^{2} =324-18x\\x^{2} +18x-324=0\\delta=324+4*1*324=5*324\\x=\frac{-18+18\sqrt{5} }{2} =-9+9\sqrt{5}\\=> BC=-9+9\sqrt{5}=9(\sqrt{5} -1)\\\\sin18=\frac{EB}{AB} =\frac{\frac{9}{2}*(-9+9\sqrt{5}) }{18} =\frac{\sqrt{5} -1}{4}[/tex]
