Answer:
sinB is correct
Step-by-step explanation:
Calculating each of cos/ sin for ∠B
cosB = [tex]\frac{adjacent}{hypotenuse}[/tex] = [tex]\frac{6}{3\sqrt{5} }[/tex] = [tex]\frac{2}{\sqrt{5} }[/tex] and
[tex]\frac{2}{\sqrt{5} }[/tex] × [tex]\frac{\sqrt{5} }{\sqrt{5} }[/tex] = [tex]\frac{2\sqrt{5} }{5}[/tex] ≠ [tex]\frac{\sqrt{5} }{5}[/tex]
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sinB = [tex]\frac{opposite}{hypotenuse}[/tex] = [tex]\frac{3}{3\sqrt{5} }[/tex] = [tex]\frac{1}{\sqrt{5} }[/tex] and
[tex]\frac{1}{\sqrt{5} }[/tex] × [tex]\frac{\sqrt{5} }{\sqrt{5} }[/tex] = [tex]\frac{\sqrt{5} }{5}[/tex]
