Question

ДОПОМОЖІТЬ БУДЬ ЛАСКА ЛАЮ 65 БАЛА

Answer 1

$\displaystyle\bf\\0 < \alpha < 90^\circ \ \ \ \Rightarrow \ \ \ Sin\alpha > 0\\\\\\Cos\alpha =\frac{\sqrt{3} }{2} \\\\\\Sin\alpha =\sqrt{1-Cos^{2} \alpha }=\sqrt{1-\Big(\frac{\sqrt{3} }{2}\Big)^{2} } =\sqrt{1-\frac{3}{4} } =\\\\\\=\sqrt{\frac{1}{4} } =\sqrt{\Big(\frac{1}{2}\Big)^{2} } =\frac{1}{2} \\\\\\Otvet \ \ . \ \ Sin\alpha =\frac{1}{2}$