Question

va rog frumos, am nevoie de ajutor​

Answer 1

Aflam AC din Pitagora

AC²=AB²+BC²

AC²=16+64=80

AC=4√5 cm

$sinC=\frac{AB}{AC} =\frac{4}{4\sqrt{5} } =\frac{\sqrt{5} }{5}$

$cosC=\frac{BC}{AC} =\frac{8}{4\sqrt{5} } =\frac{2\sqrt{5} }{5}$

$tgC=\frac{AB}{BC} =\frac{4}{8} =\frac{1}{2}$

$ctgC=\frac{BC}{AB} =2$