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. 

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. 
