How Angle AOP' is equal to (90° - θ) in the second figure? I'm learning Trigonometry right now with myself and at current I'm understanding how to find the trigonometric ratio of the angle (90°- θ) in those of θ. I'm little bit confused right now in second Figure, I didn't get how angle AOP'= (90°- θ). First figure is understandable to me because the angle BOP' is less than 90°but the second on is a little bit confusing because of the angle BOP' > 90°. Please help. Sorry If I'm asking foolish question. Thankyou in advance. 
 A: So after really looking at it some more the question is so simple it's hard. Let's start by thinking all the way back to high school geometry where you had to do triangle proofs. This is essentially the same thing, just a slightly different. 
So, let's go basic (get some paper out and follow along with me):
Let $\angle AOB$ be equal to $90^\circ$. This is true since $\angle AOB$ is a right angle. 
So if you draw this out on paper then what you're looking at is a simple right angle. Now in a different color ink on the same angle draw a line from $A$ to $O$ and let that line be known as $AO$. What it's saying is that you can rotate this line any where from $0^\circ$ ($AO$) to $90^\circ$ which is $\angle AOB$ and anything in that rotation is $\theta$. 
Now, rotate the line passed $AO$ and let any line rotated passed $AO$ be denoted as $OP$. You should now have an obtuse angle denoted as $\angle AOP$ which is $\gt 90^\circ$, so here you have $90 + \theta$. You're looking for $90 - \theta$ so we need to work backwards. 
So for this we take the line $OB$ and rotate it in the clockwise direction passed $A$, and denote any line passed $AO$ as $OP'$. Since any line going in the counter-clockwise direction passed $OB$ was considered $90 + \theta$ then logically the opposite will be $90 - \theta$.
Thus this is how you retrieve the ratio of $\angle AOP' = 90 - \theta$ because again, the rotating line generates the angle $\theta$.
Sorry for the lengthy answer but I tried to be as detailed as possible, let me know if this helps, or if you need further clarification. 
A: One thing to remember about these angles is that
we care about both the size (also called magnitude) of the angle,
we also care about an additional property that may be called
the direction, orientation, or sense of the angle.
This is unlike a lot of other plane geometry that you may have seen,
where we care only about the magnitude of the angle, not its orientation.
Rather than thinking of the angle as the shape of the wedge formed by
two rays emanating out from a common point $O$, I find it better to think
of it as a measurement of "turning" or rotation.
A positive angle is a counterclockwise rotation; negative is clockwise.
For example, if we take an arrow pointing from $O$ to $A$ and rotate it $30^\circ$ counterclockwise, the measurement of this rotation is $30^\circ$;
if we rotate it $30^\circ$ clockwise, the measurement of this rotation is $-30^\circ$. (Your book talks about rotating a line; I say "arrow" to
emphasize that the thing we are rotating points outward from $O$
to $A$ and to points beyond.)
The sum of two angles is what you get if you perform one rotation right after the other. If the rotations are in opposite directions, one of them will cancel out part of the other rotation. So if we add $80^\circ$ and
$70^\circ$ we get $150^\circ$ (all counterclockwise);
but if we add $80^\circ$ and $-70^\circ$ we get $10^\circ$ 
(that is, the $70^\circ$ clockwise rotation undoes most of the
$80^\circ$ counterclockwise rotation).
If we add $70^\circ$ and $-80^\circ$ we get $-10^\circ$,
because after rotating $70^\circ$ counterclockwise, we rotate
$80^\circ$ clockwise, which undoes all of the initial counterclockwise
rotation and then continues rotating clockwise for another $10^\circ$.
Subtracting angles just means we reverse the direction of the second angle
before adding it to the first, that is,
$80^\circ - 70^\circ = 80^\circ+ (-70^\circ) = 10^\circ$.
So $90^\circ - \theta$ means you rotate your arrow $90^\circ$ counterclockwise and then $\theta$ clockwise.
If $\theta > 90^\circ$, the clockwise rotation completely wipes out the
counterclockwise rotation and then continues to rotate clockwise
from the starting state.
For example, if $\theta = 120^\circ$, then
$90^\circ - \theta = 90^\circ - 120^\circ = -30^\circ$,
that is, if the arrow is initially pointing from $O$ to $A$ we 
turn it $90^\circ$ counterclockwise, then $90^\circ$ clockwise
(undoing the first $90^\circ$ rotation) and an
additional $30^\circ$ clockwise. The arrow ends up pointing at a
point that looks much like the point $P'$ in your second figure,
on the "clockwise" side of $OA$.
And that's what the book means when it gives $90^\circ - \theta$
as the measurement of $\angle AOP'$. Assuming $\theta$ is positive,
if you initially point the arrow toward $A$, 
rotate $90^\circ$ counterclockwise, and then rotate $\theta$ clockwise,
it ends up pointing toward $P'$. If $90^\circ < \theta < 180^\circ$
then $P'$ is in the lower right quadrant as it is in the second figure.
A: Correction:
In the second figure $\theta\approx120$, so $\angle AOP'=90-120=-30$. The direction of the measurement is important.
