reference frames - How do I choose the correct Lorentz equation for my special relativity problem?

Lets say There is a moving board moving to the right in the speed of $V = 0.8c\,\hat x$.

We know that the angle that the pole creates with the Y axis of our system is 31 degrees $(\alpha = \tan^{-1}(0.6) = 31^\circ)$.

I want to find that angle from the system of the board $\Rightarrow\alpha'$

Which Lorentz equation do I use?

(Let's say I don't want to use shortcuts like the length contraction equation, and I want to get to it by using the basic Lorentz transformation).

Answer

Let me see if I understand your question correctly: you have an object (say a rod or board) inclined at some angle $\alpha$ in some reference frame $S$, and you would like the find the angle $\alpha^\prime$ in a reference frame $S^\prime$. Since you've mentioned $\alpha^\prime$ as the angle from the rod's frame, I'm using the convention that the rest frame (i.e., the frame in which the rod is at rest) is $S^\prime$. Let me write down the Lorentz Transformations:

$$\begin{equation} \begin{aligned} &\text{(A)}\quad\Delta x^\prime = \gamma \left(\Delta x - v \Delta t\right)\\ &\text{(B)}\quad \Delta t^\prime = \gamma \left( \Delta t - \frac{v}{c^2}\Delta x\right)\\ \\ &\text{(C)}\quad\Delta x = \gamma \left(\Delta x^\prime + v \Delta t^\prime \right)\\ &\text{(D)}\quad \Delta t = \gamma \left( \Delta t^\prime + \frac{v}{c^2}\Delta x^\prime \right)\\ \end{aligned} \label{LT} \end{equation}$$

The question as to which Lorentz Transformation(s) should be used is a good one, I feel, as it's very easy to get confused and use the wrong ones. (Suddenly, you'll find lengths expanding in stead of contracting and so on! Take a look at my answer here.)

As I point out in the above answer,

For an observer sitting in $S^\prime$, since the object is at rest with respect to him, its length $L^\prime$ is simply the difference in the coordinates, irrespective of when $x_B^\prime$ and $x_A^\prime$ are measured. He could measure $x_B^\prime$, have a coffee, and then measure $x_A^\prime$ and the difference would give him the length. However, for an observer sitting in $S$, since the object is moving with respect to her, both the endpoints $x_B$ and $x_A$ need to be measured simultaneously in her frame of reference ($S$) in order for the difference to be the length $L$. (In other words, if she has a coffee between measuring $x_B$ and $x_A$, the object would have moved between measurements!) So, we have

$$L^\prime = x_B^\prime - x_A^\prime |_\text{ for any $\Delta t^\prime$}$$ $$L = x_B - x_A |_\text{ only when $\Delta t=0$}$$

Now, let's try to answer your question. You are interested in relating the angle $\alpha^\prime$ with the angle $\alpha$. From trigonometry, it's clear that

$$\tan{\alpha} = \frac{L_x}{L_y} \quad\quad \text{ and } \quad \tan{\alpha^\prime} = \frac{L_x^\prime}{L_y^\prime}.$$

Of course, since the direction of motion is only along $x$, $L_y = L_y^\prime$.

Now, all we need to do is to relate the lengths $L_x$ and $L_x^\prime$. As I've already pointed out, this means that we need to find a relation between $\Delta x$ and $\Delta x^\prime$, when $\Delta t=0$, since $L_x = \Delta x$ if (and only if) $\Delta t =0$, since the rod is moving in the frame $S$ and therefore its endpoints have to be measured simultaneously.

So, we ask ourselves, which Lorentz Transformation relates $\Delta x, \Delta x^\prime$, and $\Delta t$? The answer is, of course, (A) Remember, while $\Delta t=0$, we are saying nothing about $\Delta t^\prime$. It turns out that $\Delta t^\prime$ is not zero! This is why it isn't helpful to use (C) for example, since we'd first have to find $\Delta t^\prime$. So,

$$\begin{equation*} \begin{aligned} \Delta x^\prime &= \gamma \left(\Delta x - v \Delta t\right)\\ \Delta x^\prime|_{\Delta t = 0} &= \gamma \left(\Delta x|_{\Delta t =0} - v \Delta t|_{\Delta t = 0}\right)\\ \\ L_x^\prime &= \gamma L_x \end{aligned} \end{equation*}$$

Plugging this into our trigonometric identity, we see that

$$\tan{\alpha^\prime} = \frac{L_x^\prime}{L_y^\prime} = \frac{\gamma L_x}{L_y} = \gamma \tan{\alpha}.$$