The second formula is not valid: it should be . Solving for in this improved version of the second formula gives the first formula. So these two formulas are not fundamentally different, and therefore there is hardly any difference in their applicability to this or any other problem...

Now, since is given, you need to be able to express in this formula for in terms of .

And how do you do that? Tipp: sketch a right triangle with acute angle , adjacent leg and opposite leg . By Pyhtagoras's theorem you have that the length of the hypotenuse of this triangle is .

Therefore, by definition, we have . This is valid only for acute angles . In the general case, you need to write this as . The sign depends on the quadrant in which the point that corresponds to lies on the unit circle.

To sum it up you get:

If you are allowed to assume that is an acute angle, then you only take the case of . Now plug in the given value of ...