Please answer ASAP! If 0 is an angle in quadrant II, what is the value of sin(0)
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Answer:
Option A
Step-by-step explanation:
It's given that cosθ = [tex]-\frac{2\sqrt{5} }{5}[/tex]
Since, cosθ = [tex]\frac{\text{Adjacent side}}{\text{Hypotenuse}}[/tex] = [tex]-\frac{2\sqrt{5} }{5}[/tex]
= [tex]\frac{2}{\sqrt{5}}[/tex]
By applying Pythagoras theorem in the right triangle,
(Hypotenuse)² = (Opposite side)² + (Adjacent side)²
(√5)² = (Opposite side)² + (2)²
5 = (Opposite side)² + 4
Opposite side = √(5 - 4)
= 1
Since, sinθ = [tex]\frac{\text{Opposite side}}{\text{Hypotenuse}}[/tex]
= [tex]\frac{1}{\sqrt{5}}[/tex]
Since, sine is positive in IInd quadrant,
sinθ = [tex]\frac{1}{\sqrt{5}}[/tex]
Therefore, option A will be the answer.