Question

Find the value of sin θ, if tan θ=4; 180° < θ <270°

Answer 1

You can use the identity
$\sin(x)=\sqrt{\dfrac{\tan(x)^{2}}{1+\tan(x)^{2}}}\\\\=-\sqrt{\dfrac{4^{2}}{1+4^{2}}}\\\\=-\sqrt{\dfrac{16\cdot 17}{17\cdot 17}}=\dfrac{-4\sqrt{17}}{17}$

The sign is negative because the angle is in the 3rd quadrant.

This corresponds to ...
  a. -(4√17)/17