Options for the first box:-4/5, -3/5, 3/5, 4/5 Options for the second box:-4/3, -3/4, 3/4, 4/3

SOLUTION
We are told that
[tex]\begin{gathered} \sin \theta=\frac{3}{5} \\ i\text{n the second quadrant } \end{gathered}[/tex]In the second quadrant the required angle =
[tex]180-\theta[/tex]Also only sin is positive in the second quadrant
So
[tex]\begin{gathered} \text{sin}\theta\text{ = sin(180 -}\theta) \\ \text{sin(180 -}\theta)=\frac{3}{5} \\ \text{(180 -}\theta)=\sin ^{-1}\frac{3}{5} \\ 180-\theta=36.86989765 \\ \theta=180-36.86989765 \\ \theta=143.1301024 \end{gathered}[/tex](a)
[tex]\begin{gathered} \cos \theta=-cos(180-\theta) \\ -cos(180-\theta)=-\cos (180-143.1301024) \\ =-cos36.86989765 \\ =-\frac{4}{5} \end{gathered}[/tex](b)
[tex]\begin{gathered} \tan \theta=-\tan 36.86989765 \\ =-\frac{3}{4} \end{gathered}[/tex]