SOLUTION
From the question, the terminal angle is in quadrant 4
[tex]\begin{gathered} cos\theta=0.6 \\ cos\theta=\frac{adjacent}{hypotenuse}=\frac{6}{10} \end{gathered}[/tex]
Let's find the opposite side. from Pythagoras, we have
[tex]\begin{gathered} hyp^2=opp^2+adj^2 \\ 10^2=opp^2+6^2 \\ 100=opp^2+36 \\ opp^2=100-36 \\ opp^2=64 \\ opp=\sqrt{64}=8 \end{gathered}[/tex]
So,
[tex]\begin{gathered} sin\theta=\frac{opposite}{hypotenuse} \\ sin\theta=\frac{8}{10}=0.8 \end{gathered}[/tex]
In the 4th quadrant, only cos is positive. So, sine is negative
Hence the answer is
[tex]sin\theta=-0.8[/tex]
Option D is the answer
