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There are two parallel wires, both carry currents of $I=16.5A$ in the same direction.

The wire on the left is $B_1$ and the one on the right is $B_2$

That said I know from $B=\frac{\mu_0 I}{2\pi R}$:

$$B_1=2.7500\times10^{-5}\,\text{T}$$ $$B_2=2.5385\times10^{-5}\,\text{T}$$ Since this triangle has three diferent sides, this allows to use law of cosines. However, is there other way to solve it?

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You are right, but this rule applies only for infinite wires, otherwise use Bio-Savart-Laplace equation $$ \vec{B}=\frac{\mu_0}{4\pi} \int _{C}{\frac {Id\mathbf {l} \times \mathbf {{{r}}'} }{|\mathbf {r''} |^{2}}} $$ Where $r'$ is a location of infinitesimal element of a current-carrying wire and $r''$ - observation point (where you want to calculate)

Integrating over path of the wire gives a field, doing the same with another wire and summing them up is your answer

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You do not need the law of cosines. You know the magnitude of the fields from each wire, but in order to sum them you need their vector components. Use the lengths of the triangle to determine the direction of the two magnetic field vectors. Then you may sum the two vectors componentwise to find the net magnetic field.

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