It states that angle made by acceleration when is between 0 and 89 degrees , acceleration increases whereas in case when it is between 90 - 179 degrees, acceleration decreases.
I didn’t understand why not to include 90 and 180 degree angle ? I am not getting how to draw those components like which reach this solution ?
It states that angle made by acceleration when is between 0 and 89 degrees , acceleration increases whereas in case when it is between 90 - 179 degrees, acceleration decreases.
That's not what it says.
The textbook is talking about the instantaneous change in speed (the magnitude of velocity, which is pointing to right in the example), and about how it depends on the angle between the velocity vector and the acceleration vector.
All it's saying is this:
Remember that the direction of the acceleration, when at 90° exactly, is always perpendicular (Normal) to the direction of the velocity. And when the direction of the velocity vector changes, the direction of the acceleration vector changes with it, remaining 90°, Normal, to the velocity vector.
With that, remember that only the tangential component of the acceleration has an impact on the magnitude of the velocity.
As a perpendicular acceleration vector has no impact on the tangential speed, the magnitude of the velocity will stay constant, even though its direction will change.
In your diagram, I think you flipped around the cos and sin. if your acceleration is tangential to your velocity, $a_t = a \cdot 1$ which can only be achieved with $a \cdot \sin(0)$. Meanwhile, the normal acceleration is $a_n = a \cdot \cos(0) = a \cdot 0 = 0$.
In the same way, normal acceleration has no impact on the magnitude of the velocity because $a_t = a \cdot \sin(90) =0$.
If the force (i.e. acceleration) is perpendicular to the particle's velocity, the absolute value of the velocity will not change but only its direction. Think about a pearl going around on a ring. Its position can be described by a vector pointing from the center of the center of the circle to the position of the pearl on the ring. Its velocity is always perpendicular to that vector, and indeed the distance of the pearl form the center always has the same magnitude.
Therefore, $90^\circ$ is a special case. That works for your text as well: as long as $0^\circ \le \theta < 90^\circ$, there is an increase in velocity, and as soon as $90^\circ < \theta \le 180^\circ$, there is a decrease. All these functions are continuous, so if the change of the magnitude of the velocity changes sign when going from above $90^\circ$ to below $90^\circ$, it must be that this change is exactly $0$ at $90^\circ$.
The $180^\circ$ angle is also included, I think you misread the $\le$ and $<$ signs there.
Edit: Also, because you ask why we don't consider the angle of $90^\circ$: we do. It's just neither an angle at which the absolut velocity dereases nor one at which it increases, so it doesn't appear in the list. This is even said in the note right below the sentence you were asking about.
Edit 2: The relation shown with vectors. We want to know how the absolute value of the velocity changes. Take its square, $|v(t)|^2 = \mathbf{v}(t) \cdot \mathbf{v}(t)$, and calculate the time derivative $$\frac{\textrm{d}}{\textrm{d}t} \mathbf{v}(t) \cdot \mathbf{v}(t) = 2 \mathbf{v}(t) \cdot \left[\frac{\textrm{d}}{\textrm{d}t}{\mathbf{v}}(t)\right] = 2 \mathbf{v}(t) \cdot \mathbf{a}(t).$$ If $\mathbf{v}(t) \cdot \mathbf{a}(t) > 0$, then the absolute value increases, and this is equivalent to $0 \le \theta < 90^\circ$. If $\mathbf{v}(t) \cdot \mathbf{a}(t) < 0$, then the absolute value decreases, and this is equivalent to $90^\circ < \theta \le 180^\circ$. If, finally, $\mathbf{v}(t) \cdot \mathbf{a}(t) = 0$, then the modulus $|v(t)|$ does not change in time, the absolute value of the velocity remains constant. This corresponds to $\theta = 90^\circ$.
Edit 3: I assume from the further questions that you may have problems with a common identity. Take any two vectors $\mathbb{u}$ and $\mathbb{w}$. The angle $\theta_{uw}$ enclosed by these vectors is defined via $$\mathbb{u} \cdot \mathbb{w} = |\mathbb{u}||\mathbb{w}|\cos(\theta_{uw}).$$ This definition can be made for any type of vectors, but especially from three-dimensional real vectors like those describing velocity and acceleration. The angle $\theta$ in your example is the angle formed between $\mathbb{v}(t)$ and $\mathbb{a}(t)$. If you struggle with these concepts, you should pick up any introductory math for physicists book and get familiar with them.
