Question

4. Qual o valor de N= 3.cos 225°- 4.sen 330° +2.tg 150° ?​

Answer 1

N = 3 . cos(225°) - 4 . sin(330°) + 2 . tan(150°)

N = 3 . (- √2/2) - 4 . (- 1/2) + 2 . (- √3/3)

N = - 3 . √2/2 + 4 . 1/2 - 2 . √3/3 : (2)

N = - 3 . √2/2 + 2 - 2 . √3/3

N = - 3√2/2 + 2 - 2√3/3

N ≊ - 2,12 + 2 - 1,15

N ≊ - 3,27 + 2

N ≊ - 1,27

atte. yrz