Question

Daca ma poate ajuta cineva cu problema de jos pls

Answer 1

Salut,

Conform figurii avem așa:

$tg30^{\circ}=\dfrac{x}{4\sqrt3}\Leftrightarrow\dfrac{\sqrt3}3=\dfrac{x}{4\sqrt3}\Leftrightarrow \dfrac{1}{\sqrt 3}=\dfrac{x}{4\sqrt3}\Rightarrow x=4\ cm.$

Apoi:

$sin30^{\circ}=\dfrac{x}y\Leftrightarrow\dfrac{1}2=\dfrac{4}y,\ deci\ y=8\ cm.$

Ai înțeles ?

Green eyes.