Let's take a closer look at our vector:
Now, notice that the angle a and the 130° angle given in the plot are a linear pair. This way,
![\begin{gathered} a+130=180\rightarrow a=180-130 \\ \Rightarrow a=50 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/30qlmrwqt30ze218ko9zteguea4a5jt1ir.png)
Therefore, our graph would be:
Since this is a right triangle, we'll have that:
![\cos 50=\frac{Wx_{}}{W}\rightarrow Wx=W\cos 50](https://img.qammunity.org/2023/formulas/mathematics/college/3lrn68cicxlhwhwh3sjydnxln23vqdag8k.png)
And that:
![\sin 50=(Wy)/(W)\rightarrow Wy=W\sin 50](https://img.qammunity.org/2023/formulas/mathematics/college/s1k6ecmvcn2rnnbhual3zcc1wm06f4d6fy.png)
Notice that the horizontal component is going to the left (negative) and that the vertical component is going down (negative). This way, we'll have that:
![\begin{gathered} Wx=-W\cos 50 \\ Wy=-W\sin 50 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/bfrguikxypbi17rfxwnpe6ow0ie3uih2b6.png)
Plugging in the magnitude of the vector (19),
![\begin{gathered} Wx=-19\cos 50\rightarrow Wx=-12.2 \\ Wy=-19\sin 50\rightarrow Wy=-14.6 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/9s6ixmcmw14hksq1yxeb9nse9cs6921mpe.png)
This way, we can conclude that the horizontal component is -12.2 and that the vertical component is -14.6