Answer: II, positive, [tex]\frac{\pi}{3}[/tex], C, [tex]\frac{\sqrt{3}}{2}[/tex]
Step-by-step explanation:
a) [tex]\frac{2\pi}{3}[/tex] = 120° which is located in Quadrant II
b) In Quadrant II, cos is negative and sin is positive
c) In order to reach the x-axis, it would need to go to 180°. 180 - 120 = 60° ... or ... it would need to go to π. π - [tex]\frac{2\pi}{3}[/tex] = [tex]\frac{\pi}{3}[/tex]
d) sin [tex]\frac{2\pi}{3}[/tex] = [tex]\frac{\sqrt{3}}{2}[/tex]. What other angle on the Unit Circle has that value for sin? [tex]\frac{\pi}{6}[/tex]
e) sin [tex]\frac{2\pi}{3}[/tex] = [tex]\frac{\sqrt{3}}{2}[/tex]
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Step-by-step explanation:
(16, -18) are the legs of the triangle. Use Pythagorean Theorem to find the hypotenuse. 16² + (-18)² = c² ⇒ 580 = c² ⇒ [tex]2\sqrt{145}[/tex] = c
adjacent = 16, opposite = -18, hypotenuse = [tex]2\sqrt{145}[/tex]
Answers:
sin = [tex]\frac{opposite}{hypotenuse}[/tex] = [tex]\frac{-18}{2\sqrt{145}}[/tex] = [tex]\frac{-9}{\sqrt{145}} * \frac{\sqrt{145}}{\sqrt{145}}[/tex] = [tex]\frac{-9\sqrt{145}} {145}[/tex]
cos = [tex]\frac{adjacent}{hypotenuse}[/tex] = [tex]\frac{16}{2\sqrt{145}}[/tex] = [tex]\frac{8}{\sqrt{145}} * \frac{\sqrt{145}}{\sqrt{145}}[/tex] = [tex]\frac{8\sqrt{145}} {145}[/tex]
tan = [tex]\frac{opposite}{adjacent}[/tex] = [tex]\frac{-18}{16}[/tex] = [tex]-\frac{9}{8}[/tex]
csc = [tex]\frac{hypotenuse}{opposite}[/tex] = [tex]\frac{2\sqrt{145}}{-18}[/tex] = [tex]\frac{\sqrt{145}}{-9}[/tex] = [tex]-\frac{\sqrt{145}} {9}[/tex]
cos = [tex]\frac{hypotenuse}{adjacent}[/tex] = [tex]\frac{2\sqrt{145}}{16}[/tex] = [tex]\frac{\sqrt{145}}{8}[/tex]
cot = [tex]\frac{adjacent}{opposite}[/tex] = [tex]\frac{16}{-18}[/tex] = [tex]-\frac{8}{9}[/tex]
