Trigonometry & circle math I tried to solve this Trigonometry question, but I do not know how to solve. I read that the 
circle has radius 1 and center at (0.0) as the unit circle is plotted in the coordinate 
system. I should record the angles $v$ and $w$ in unit circle so that the following are met:
$v$ is an acute angle with $\sin(v) = 0.9$ and w is an obtuse angle with $\sin(w) = 0.9$
You can see the circle here
i don't know how to solve it, but is it something like : $(\sin(0.9) * 100) - \pi = 75.1910983$ ? please help me out, give me examples and information
 A: I think you simply need to find the points on the unit circle with $y$-coordinate $.9$ (recall $\sin\theta$ is the $y$ coordinate of the appropriate point on the unit circle).    
For the acute angle, $v$,  with $\sin v=.9$,  $v$ is in the first quadrant. The $y$ coordinate is $.9$, and using the Pythagorean Theorem, the $x$-coordinate is $x=\sqrt{1-.9^2}=\sqrt{19/100}={1\over10}\sqrt{19}$.

The obtuse angle is in the second quadrant with $y$-coordinate $.9$. You can solve for the $x$-coordinate as above. 
I'm not sure that you need explicitly find the measure of the angle; but if so, take the inverse sin of the angles. 
A: Essentially, you're looking for an angle $v$ that satisfies: $$ \sin(v) = 0.9 \;\;\; 0\leq v \leq \frac{\pi}{2}$$ (angles in rad)  an angle $w$ that satisfies: $$ \sin(w) = 0.9 \;\;\; \frac{\pi}{2}\leq w\leq \pi$$  The value for $v$ follows easily from the first equation; just take the inverse sine of both sides.  For the second equation, recall that $\sin(\pi - \theta) = \sin(\theta)$.  Therefore, $w = \pi - v$
Note that this is a purely algebraic approach, and ignores the unit circle, which may well be the point of the exercise.  I refer you to David's answer for a better explanation of the geometric approach.
