Question

(UFPR) O número
N= (3.cos180° - 4.sen120° + 2.tg135°) / (6.sen^2 45°)
Pertence ao intervalo
A) ]-4, -3[
B) [-3,-2[
C) [-2,-1]
D) [-1, 0]

Answer 1

$N = \frac{3cos\: 180 - 4sen\: 120 + 2tg\: 135 }{6sen^2\: 45} \\ \\ N = \frac{3(- 1) - 4 \frac{ \sqrt{3} }{2} + 2.1) }{6( \frac{ \sqrt{2} }{2})^2 } \\ \\ N = \frac{- 3 - 2 \sqrt{3} + 2 }{6 \frac{2}{4} } \\ \\ N = \frac{- 2 \sqrt{3} - 1 }{3}$

sen 120° = sen 60° = √3/2
tg 135° = - tg 45° = - (- 1) = 1