Question

Encontre a 1º determinação, ou seja, o menor valor não negativo côngruo ao arco de:
A)780°
B)1140°
C)-400°
D)15/2pi rad
E)10/3 pi rad
F)9/2pi rad

Answer 1

a)

$\mathsf{780^{\circ}=2\times360^{\circ} +20^{\circ}}\\\mathsf{\alpha=20^{\circ}+2k\pi,k\in\mathbb{Z}}$

b)

$\mathsf{1140^{\circ}=3\times360^{\circ}+60^{\circ}}\\\mathsf{\alpha=60^{\circ}+2k\pi,k\in\mathbb{Z}}$

c)

$\mathsf{-400^{\circ}=-1\times360^{\circ}-40^{\circ}}\\\mathsf{360-40=320^{\circ}}\\\mathsf{\alpha=320^{\circ}+2k\pi,k\in\mathbb{Z}}$

d)

$\mathsf{\dfrac{15\pi}{2}=\dfrac{12\pi}{2}+\dfrac{3\pi}{2}=6\pi+\dfrac{3\pi}{2}}\\\mathsf{\dfrac{15\pi}{2}=3\times2\pi+\dfrac{3\pi}{2}}\\\mathsf{\alpha=\dfrac{3\pi}{2}+2k\pi,k\in\mathbb{Z}}$

e)

$\mathsf{\dfrac{10\pi}{3}=\dfrac{6\pi}{3}+\dfrac{4\pi}{3}}\\\mathsf{\dfrac{10\pi}{3}=1\times2\pi+\dfrac{4\pi}{3}}\\\mathsf{\alpha= \dfrac{4\pi}{3}+2k\pi,k\in\mathbb{Z}}$

f)

$\mathsf{\dfrac{9\pi}{2}=\dfrac{8\pi}{2}+\dfrac{\pi}{2}=4\pi+ \dfrac{\pi}{2}}\\\mathsf{ \dfrac{9\pi}{2} = 2\times2\pi+ \dfrac{\pi}{2}}\\\mathsf{\alpha=\dfrac{\pi}{2}+2k\pi,k\in\mathbb{Z}}$