Răspuns:
Din Th. lui Pitagora BC² = AC² + MP²
BC² = 3² + (3√2)²
BC² = 9 + 9*2
BC² = 9 + 18 ⇒ BC = √27 ⇒ BC = 3√3
AD =[tex]\frac{cateta 1 * cateta2}{ipotenuza} =[/tex] [tex]\frac{AC * AB}{BC} =[/tex] [tex]\frac{3 * 3\sqrt{2}}{3\sqrt{3} } =[/tex] [tex]\frac{3\sqrt{2} }{\sqrt{3} }[/tex]
cos B = [tex]\frac{cateta alaturata}{ipotenuza} = \frac{3\sqrt{2} }{3\sqrt{3} } = \frac{\sqrt{2} }{\sqrt{3} }[/tex]
sin B = [tex]\frac{cateta opusa}{ipotenuza} = \frac{AC}{BC} = \frac{3}{3\sqrt{3} } = \frac{1}{\sqrt{3} }[/tex]
tg B = [tex]\frac{cateta opusa}{catega alaturata} = \frac{AC}{AB} = \frac{3}{3\sqrt{2} } = \frac{1}{\sqrt{2} }[/tex]
ctg B = [tex]\frac{cateta alaturata}{cateta opusa} = \frac{AB}{AC} = \frac{3\sqrt{2} }{3} = \frac{\sqrt{2} }{1} = \sqrt{2}[/tex]
Explicație pas cu pas:
