Bună!
Δ₉₀°ABC: m(∡ABC)=60° ⇒ m(∡ACB)=30°
Metoda 1
sinB=[tex]\frac{AC}{BC}[/tex] ⇔ [tex]\frac{\sqrt{3} }{2} =\frac{10}{BC} => BC=^{\sqrt{3} )} \frac{2*10}{\sqrt{3} } => BC=\frac{20\sqrt{3} }{3}[/tex]
T∡30°⇒ [tex]AB=\frac{BC}{2}[/tex] ⇒ [tex]AB=\frac{20\sqrt{3} }{3} *\frac{1}{2} =\frac{10\sqrt{3} }{3}[/tex]
Metoda 2
notăm: BC=x , AB=[tex]\frac{x}{2}[/tex]
T.P⇒ [tex]10^{2} +(\frac{x}{2} )^{2} =x^{2} <=> 100+\frac{x^{2} }{4} =x^{2}/*4 <=> 400+x^{2} =4x^{2} <=> x^{2} - 4x^{2}=-400 <=> -3x^{2} =-400 => x^{2} =\frac{400}{3} => x= +-\sqrt{\frac{400}{3} } => x=+-\frac{20\sqrt{3} }{3} => x=\frac{20\sqrt{3} }{3}[/tex]
→ Laturile pot lua doar valori pozitive
AB=[tex]\frac{x}{2} = \frac{20\sqrt{3} }{3} *\frac{1}{2} =\frac{10\sqrt{3} }{3}[/tex]
