Answer:
C
Step-by-step explanation:
Given
tanθ = [tex]\frac{1}{2}[/tex] = [tex]\frac{opposite}{adjacent}[/tex]
This is a right triangle with legs 1 and 2
Using Pythagoras' identity
hypotenuse² = 1² + 2² = 1 + 4 = 5 ( take square root of both sides )
hypotenuse = [tex]\sqrt{5}[/tex]
θ is in the 3rd quadrant where sinθ < 0 , then
sinθ = [tex]\frac{opposite}{hypotenuse}[/tex] = - [tex]\frac{1}{\sqrt{5} }[/tex] ← rationalise the denominator
= - [tex]\frac{1}{\sqrt{5} }[/tex] × [tex]\frac{\sqrt{5} }{\sqrt{5} }[/tex]
= - [tex]\frac{\sqrt{5} }{5}[/tex] → C
