Question

(fgv-SP) na figura, x e y valem respectivamente:

Answer 1

$\tan(60) = \frac{9}{x} \\ \sqrt{3} = \frac{9}{x} \\ x \sqrt{3} = 9 \\ x = \frac{9}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } \\ \frac{9 \times \sqrt{3} }{3} \\ 3 \sqrt{ 3} \\ \\ \\ \tan(45) = \frac{9}{3 \sqrt{3} + y } \\ 1 = \frac{9}{ 3\sqrt{3} + y} \\ 9 = 3 \sqrt{3 } + y \\ - 3 \sqrt{3} + 9 = y \\ 3(3 - \sqrt{3} )$
letra A