Ropes 3 m and 5 m in length are fastened to a holiday decoration that

Ropes 3 m and 5 m in length are fastened to a holiday decoration that is suspended over a town square. The decoration has a mass of 5 kg. The ropes, fastened at different heights, make angles of 52° and 40° with the horizontal. Find the tension in each wire and the magnitude of each tension.
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Toni Scott
There are two things to notice. The first is that the length of the rope is not important in terms of finding the tension vectors. The second is that the mass of the decoration is 5 kg. That means that the force is in the negative j direction. If we set the problem up in the same way as in the example, then ${T}_{1}$ is then tension in the left rope and ${T}_{2}$ is the tension in the tight rope. We see that
${T}_{1}=-|{T}_{1}|\mathrm{cos}52i+|{T}_{1}|\mathrm{sin}52j$
${T}_{2}=|{T}_{2}|\mathrm{cos}40i+|{T}_{2}|\mathrm{sin}40j$
$w=-49j$
Since the decoration is not moving, we have
$0={T}_{1}+{T}_{2}+w$
$=-|{T}_{1}|\mathrm{cos}52i+|{T}_{1}|\mathrm{sin}52j+|{T}_{2}|\mathrm{cos}40i+|{T}_{2}|\mathrm{sin}40j-49j$
$=\left(-|{T}_{1}|\mathrm{cos}52+|{T}_{2}|\mathrm{cos}40\right)i+\left(|{T}_{1}|\mathrm{sin}52+|{T}_{2}|\mathrm{sin}40-49\right)j$
This gives us a system of equations:
$-|{T}_{1}|\mathrm{cos}52+|{T}_{2}|\mathrm{cos}40=0$
$|{T}_{1}|\mathrm{sin}52+|{T}_{2}|\mathrm{sin}40-49=0$
Solving for $|{T}_{2}|$ in the first equation and solving for $|{T}_{1}|$, we get
$|{T}_{2}|=\frac{|{T}_{1}|\mathrm{cos}52}{\mathrm{cos}40}$
Plugging this into the second equation and solving for $|{T}_{1}|$, we get
$|{T}_{1}|=\frac{49}{\mathrm{sin}52+\mathrm{cos}52\mathrm{tan}40}\approx 37.6$
We then get that
$|{T}_{2}|=\frac{|{T}_{1}|\mathrm{cos}52}{\mathrm{cos}40}\approx 30.2$
Therefore we get
${T}_{1}=-23.1i+29.6j$
${T}_{2}=23.1i+19.4j$