Decide tan v if sin v = 1/√3 and if π/2 < v < π

sin v = 1/√3 and if π/2 < v < π       find tan v

It is in the second quadrant so tan will be negative.

Now I can just 'pretend' it is in the first quadrant for  all the working and then just add the minus sign at the end.

\(sin\; v=\frac{1}{\sqrt3}=\frac{opp}{hyp}\\ \text{Using pythagoras, the adjacent side must be }\sqrt{3-1}=\sqrt2\\ so\\ tan\;v=-\frac{opp}{adj}=-\frac{1}{\sqrt2}\)

Here is a represention of it on a number plane.