So I have a math problem that involves one vector from $(0,0)$ and another one which I do not know how to write; it starts at 197.7 on the $x$-axis and goes up to $y=30$, I need to add this vector to a different vector but I do not quite know how to write it in rectangular form.
Thanks!
You seem to be confused about the nature of vectors, or at least the terminology.
Let's consider three location vectors, each specifying a point on the 2D plane: $$\begin{aligned} O &= (0, 0) = \left [ \begin{matrix} 0 \\ 0 \end{matrix} \right ] \\ A &= (197.7, 30) = \left [ \begin{matrix} 197.7 \\ 30 \end{matrix} \right ] \\ B &= (40, 20) = \left [ \begin{matrix} 40 \\ 20 \end{matrix} \right ] \end{aligned}$$
We can also define spatial vectors (also called Euclidean vectors), between any two points. For example, $$\begin{aligned} \vec{a} &= \overline{OA} = (197.7-0, 30-0) = (197.7, 30) = \left [ \begin{matrix} 197.7 - 0 \\ 30 - 0 \end{matrix} \right ] = \left [ \begin{matrix} 197.7 \\ 30 \end{matrix} \right ] \\ \vec{b} &= \overline{BA} = ( 197.7-40, 30-20) = ( 157.7, 10 ) = \left [ \begin{matrix} 197.7 - 40 \\ 30 - 20 \end{matrix} \right ] = \left [ \begin{matrix} 157.7 \\ 10 \end{matrix} \right ] \end{aligned}$$
All vectors above use Cartesian coordinates.
It is obviously possible to use other coordinate systems; polar coordinates are one possible choice: $$\begin{cases} x = r \cos \varphi \\ y = r \sin \varphi \end{cases} \quad \iff \quad \begin{cases} r = \sqrt{x^2 + y^2} \\ \varphi = \operatorname{atan2}(y, x) \end{cases}$$ where $\operatorname{atan2}(y,x)$ is like $\arctan(y/x)$, except that $\operatorname{atan2}$ includes the correct quadrant also (that depends on the signs of $x$ and $y$). In polar coordinates, $r$ specifies the distance, and $\varphi$ the direction.
In these polar coordinates, $\varphi = 0°$ on the positive $x$ axis, $\varphi = 90°$ on the positive $y$ axis, $\varphi = 180°$ on the negative $x$ axis, and $\varphi = 270°$ on the negative $y$ axis. (Assuming a right-handed coordinate system with positive $x$ axis right and positive $y$ axis up, this means $\varphi$ is counterclockwise from positive $x$ axis.)
If we look at OP's question, only one point, $(197.7, 30)$ is well defined. Let's say we call that $A = (197.7, 30)$.
OP mentions another vector starting at $x = 197.7$, and "going up to 30". This is so Spinal Tappy I can only guess at what is meant by it. Nevertheless, I shall try.
Let's assume the point (or location vector) OP talks about ("starts at x = 197.7 and goes up 30") is $A = ( 197.7 , 30 )$.
OP also mentions "a different vector". Let's say this is $\vec{p} = ( x , y )$. Then,
$$A + \vec{p} = \left [ \begin{matrix} 197.7 + x \\ 30 + y \end{matrix} \right ] = ( 197.7 + x , 30 + y )$$
In English, one might call $A + \vec{p}$ "$A$ translated by $\vec{p}$" or "$\vec{p}$ relative to point $A$", specifying the location where $A$ would be if origin was at $\vec{p}$, or where $\vec{p}$ would be if origin was at $A$.
Mathematically, it does not matter whether a vector is a location or spatial vector; the rules and operations are always the same. It only matters to us humans, and not very much even to us because the correct interpretation is always found in the context anyway; and that's why we use the name vector for both.
For example, point or location vector $B = (40, 20)$ is exactly the same as spatial vector $\vec{b}$ having length $r = \sqrt{40^2 + 20^2} \approx 44.721$ and direction $\varphi = \operatorname{atan2}(20, 40) \approx 26.565°$, or Cartesian coordinate components $x = 40$ and $y = 20$. In any case, $B = \vec{b}$.
If someone just declares a vector $B = (40, 20)$ or $\vec{b} = (40, 20)$, it does not matter whether they define it as being a location or spatial vector: we can use/interpret it as either, based on the demands of the situation. The math does not care; all that is relevant is that it is a vector in Cartesian coordinates with components $x = 40$ and $y = 20$.