You can use the identity
[tex]\sin(x)=\sqrt{\dfrac{\tan(x)^{2}}{1+\tan(x)^{2}}}\\\\=-\sqrt{\dfrac{4^{2}}{1+4^{2}}}\\\\=-\sqrt{\dfrac{16\cdot 17}{17\cdot 17}}=\dfrac{-4\sqrt{17}}{17}[/tex]
The sign is negative because the angle is in the 3rd quadrant.
This corresponds to ...
a. -(4√17)/17
