Question

(Sen 45°)^-1 + cos 60°
--------------------------------
cos 30°​

Answer 1

Por triángulos notables se sabe que:

Sen45 = $\sqrt{2}/2$

Cos60 = 0.5

Cos 30 = $\sqrt{3}/2$

$((sen45)^{-1} + cos60)/cos30\\\\(\frac{2}{\sqrt{2}} + 0.5)/(\sqrt{3}/2) \\\\1.914/0.866\\\\2.21$