The title might be a bit confusing but I'll do my best to explain this. So this image is given:
The coordinates of vertex A are (3,2). We also know that [AD] and [BC] measure 1 unit each and [AB] &[DC] measure 1.5 units each. The gun of this "tank" is aligned with the tank. The slope of the direction of the gun is +0.6.
All this information was given. So far I was able to find the value of alpha, which is arctan(0.6). But I have no idea how to use this information to actually find the values of B, D and C.
I have been studying trigonometry for like 3 weeks and this seems pretty hard to solve. So far i know how the unit circle works and how to find the sine,cosine and tangent using the unit circle but I am not sure if the unit circle is needed here. I also know about triangles, but I can't see any connection with this exercise. I would appreciate any help I could get.
Let's start by taking the vertex A as our origin, and let's assume that the tank is aiming directly along the $x$-axis, so that the slope and the angle are zero. It's easy see now that, with A being our origin, the coordinates of $B, C$ and $D$ are $$ \begin{cases} B &= (1.5,0) \\ C &= (1.5,-1) \\ D &= (0,-1) \end{cases} $$ Now, if we start increasing the angle, what happens? Well, these point can be thought of vectors that start rotating about the origin. In general, when a vector $\vec v$ is rotated about the origin by the angle $\alpha$ (positive direction is anti-clockwise), the new coordinates can be calculated with the matrix multiplication $$ \vec v_{\text{new}} = \left[ \begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array} \right] \vec v $$ Now we only have to apply this to coordinates for $B, C$ and $D$! $$ B_{\text{new}} = \left[ \begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array} \right] \left[ \begin{array}{cc} 1.5 \\ 0 \end{array} \right] = \left[ \begin{array}{cc} 1.5 \cos \alpha \\ 1.5 \sin \alpha \end{array} \right] $$ $$ C_{\text{new}} = \left[ \begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array} \right] \left[ \begin{array}{cc} 1.5 \\ -1.0 \end{array} \right] = \left[ \begin{array}{cc} 1.5 \cos \alpha - 1.0 \sin \alpha \\ 1.5 \sin \alpha - 1.0 \cos \alpha \end{array} \right] $$ $$ C_{\text{new}} = \left[ \begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array} \right] \left[ \begin{array}{cc} 0 \\ -1.0 \end{array} \right] = \left[ \begin{array}{cc} 1.0 \sin \alpha \\ -1.0 \cos \alpha \end{array} \right] $$ Phew, that's the heavy lifting. Now we only need to add the coordinates of $A$ to each one of these to get the final coordinates. I think you can probably do it.
A point about numerics: Which one is given, the slope or the angle? If it's the angle, then it's better to use the equations above. However, if you have the slope, it's perhaps not optimal to first calculate the arctangent of the slope, but rather try to directly calculate the sine and cosine. You can do it knowing that $$ \text{slope} = \tan \alpha = \frac{\sin \alpha}{ \cos \alpha} $$