Question

Am nevoie de acest Exercițiu ,va rog

Answer 1

$\it sin150^0 =sin(180^o-150^o) =sin30^o=\dfrac{1}{2} \\ \\ \\ sin120^0 =sin(180^o-120^o) =sin60^o=\dfrac{\sqrt3}{2} \\ \\ \\ E= \dfrac{\dfrac{1}{2}-\dfrac{\sqrt3}{2}}{\dfrac{1}{2}+\dfrac{\sqrt3}{2}} =\dfrac{\dfrac{1-\sqrt3}{2}}{\dfrac{1+\sqrt3}{2}} =\dfrac{1-\sqrt3}{1+\sqrt3} =\dfrac{^{\sqrt3-1)}1-\sqrt3}{\ \ \sqrt3 +1} =\dfrac{\sqrt3-3-1+\sqrt3}{(\sqrt3)^2-1^2}= \\ \\ \\ = \dfrac{2\sqrt3-4}{3-1} =\dfrac{2(\sqrt3-2)}{2} =\sqrt3-2$