Answer:
sin θ = [tex]\frac{\sqrt{6} }{3}[/tex]
Step-by-step explanation:
Given that,
cos θ = [tex]\frac{\sqrt{3} }{3}[/tex]
From the trigonometric functions,
cos θ = [tex]\frac{adjacent}{hypotenus}[/tex]
⇒ adjacent = [tex]\sqrt{3}[/tex], and the hypotenuse = 3
Let the opposite side be represented by x, applying the Pythagoras theorem we have;
[tex]/hyp/^{2}[/tex] = [tex]/adj/^{2}[/tex] + [tex]/opp/^{2}[/tex]
[tex]/3/^{2}[/tex] = [tex](\sqrt{3} )^{2}[/tex] + [tex]x^{2}[/tex]
9 = 3 + [tex]x^{2}[/tex]
[tex]x^{2}[/tex] = 9 - 3
= 6
x = [tex]\sqrt{6}[/tex]
Thus, opposite side = [tex]\sqrt{6}[/tex]
So that,
sin θ = [tex]\frac{opposite}{hypotenuse}[/tex]
= [tex]\frac{\sqrt{6} }{3}[/tex]
Therefore,
sin θ = [tex]\frac{\sqrt{6} }{3}[/tex]
