Here, a Bob is hanging inside a block which is moving with acceleration $7.5m/s^2$.
Then, it has been solved that if the block moves with this acceleration. The angle subtended by the Bob inside the block is 37 degree. Now, my Q is if the Bob is accelerating, not moving with constant velocity. Then , it’s angle would keep on changing. There is no one value of theta where it would remain constant. If yes , then how is it that for this Q we were able to find a value for Theta.
If you obtain the equation of motion you get:
$$\ddot \varphi+\frac{g\,\sin(\varphi)-a\,\cos(\varphi)}{l}=0\tag 1$$
where a is the acceleration.
for a steady state $~\ddot \varphi=0~$ you obtain
$$a=\tan(\varphi_0)\,g$$
for non steady state you have to solve first the differential equation and get $~\varphi(t)$, again for $~\ddot \varphi(t)=0~$ you get:
$$a=\tan(\varphi(t)\bigg|_{\ddot \varphi=0})\,g$$
simulation results:
data : $~\varphi_0=0.1~,a=2~,l=1~,g=10$
you can see that for $~\ddot\varphi(t)=0~,\frac{a}{g}=0.2=~$ const.
In the frame of reference moving with the bob, the bob experiences three forces: gravity, the force from tension, and the fictitious force $-ma$ from the acceleration $a$. The fictitious force arises because we are observing the bob from an accelerating (non-inertial) frame of referene. The bob swings as shown, increasing the angle $\theta$, until the component of tension to the right equals the fictitious force to the left. At that time, these two forces cancel and the bob remains at a fixed final angle $\theta$. The component of tension upward counters the force of gravity downward. So, at the fixed angle $\theta$, the bob is in equilibrium and $\theta$ remains fixed. At this time, $Tsin \theta = ma$, and $Tcos\theta = mg$ as your figure shows.
It is tension that counters the fictitious force and keeps the bob fixed in the moving reference frame.
In the moving reference frame the equilibrium position of the bob is at the angle $tan\theta = a/g$ as indicated in your picture. (With no acceleration of the block, $a = 0$ and $\theta = 0$; the bob hangs vertically.) If you give the bob a slight tap, it acts as a pendulum and oscillates about the equilibrium angle $\theta$ which is not zero when the block is accelerating. See Time period of a simple pendulum in an accelerated frame on this exchange.
If I get it right, you are getting confused over the fact that when viewed from earth, we will see the bob accelerating forward (alongside the box), which means that the bob will have a different velocity from before at each instant, so how can we set an angle of the string with the vertical, at which the bob will move?
The reason is that the bob is not the only one that is accelerating from the ground frame. The box to which it is tied is also accelerating. What's more, the horizontal component of the box is the same as that of the bob. So, for every distance that the bob travels, the roof will also travel the same distance forward, keeping the angle constant.
This can be seen in how we make both the bob and the box come to a stop on changing the frame of reference to the box. In this F.O.R. , we (the observer) move the same distance that the bob would have moved, if it were viewed from F.O.R. taken as earth, making the bob move an effective of $0$ distance in this F.O.R. Hence, the angle it makes with the vertical is constant as the bob is effectively at rest W.R.T. the observer in the F.O.R. of the box.
