Let us define $f:
\begin{bmatrix}
a_2\\
\phi_2
\end{bmatrix}\mapsto
\begin{bmatrix}
f_1(a_2, \phi_2)\\
f_2(a_2, \phi_2)
\end{bmatrix} =
\begin{bmatrix}
\sqrt{a_1^{2}+a_2^{2}+a_1 a_2 cos(\phi_1 - \phi_2)}-A\\
\frac{a_1 sin(\phi_1)+a_2 sin(\phi_2)}{a_1 cos(\phi_1)+a_2 cos(\phi_2)}-tan(\phi)
\end{bmatrix}$
where $a_1, \phi_1, A, \phi$ are supposed to be known.
Then I'd go for numerical tools such as the Newton method.
First you need find a way to compute your gradient, either analytically or numerically compute the gradient matrix, here
$G(a_2, \phi_2) =
\begin{bmatrix}
\frac{\partial f_1}{\partial a_2}(a_2, \phi_2) & \frac{\partial f_1}{\partial \phi_2}(a_2, \phi_2)\\
\frac{\partial f_2}{\partial a_2}(a_2, \phi_2) & \frac{\partial f_2}{\partial \phi_2}(a_2, \phi_2)\\
\end{bmatrix}$
Then you enter the algorithm. You need to find an initial seed as close as you can to the solution ($X_0 = \begin{bmatrix}a_1\\ \phi_1\end{bmatrix}$ in our case for instance) and do in a loop :
$X_{k+1} = X_{k} - G(X_k)^{-1}f(X_k)$
and if you just want the result, this might do the trick for you.
EDIT :
apparently you want to go full analytical. Let us use the second of your equations. We have, after some regrouping:
$$a_1(tan(\phi)cos(\phi_1)-sin(\phi_1)) = a_2(sin(\phi_2)-tan(\phi)cos(\phi_2))$$
by remarking that:
$$sin(\phi_2)-tan(\phi)cos(\phi_2) = -\sqrt{1+tan^{2}(\phi)}[\frac{tan(\phi)}{\sqrt{1+tan^{2}(\phi)}}cos(\phi_2)-\frac{1}{\sqrt{1+tan^{2}(\phi)}}sin(\phi_2)]$$
we can, by setting $\theta = arctan(1/tan(\phi))$ get $cos(\theta) = \frac{tan(\phi)}{\sqrt{1+tan^{2}(\phi)}}$ and $sin(\theta) = \frac{1}{\sqrt{1+tan^{2}(\phi)}}$. Then by using the $cos(\phi_2+\theta)$ development formula backwards we get :
$$a_2cos(\phi_2+\theta) = -\frac{a_1(tan(\phi)cos(\phi_1)-sin(\phi_1))}{\sqrt{1+tan^{2}(\phi)}}$$. We will call the right hand factor $m$ for the sake of brevity. Note that is does not depend on $\phi_2$ nor $a_2$. So we have
$$a_2cos(\phi_2+\theta) = m \tag{1}\label{eq1}$$
Note how $m$ does not depend on $\phi_2$ nor $a_2$ (that was the point).
The other equation of your post can be squared and put under the form :
$$a_2^{2}+2a_1a_2cos(\phi_2-\phi_1) = A^{2}-a_1^{2}$$
Now if we develop the cosine and use \eqref{eq1}, we get
$$2a_1a_2cos(\phi_2-\phi_1)=2a_1a_2cos((\phi_2+\theta)-(\phi_1+\theta))$$ $$=2a_1a_2[cos(\phi_2+\theta)cos(\phi_1+\theta)+sin(\phi_2+\theta)sin(\phi_1+\theta)]$$ $$=2a_1mcos(\phi_1+\theta)+[2a_1sin(\phi_1+\theta)]$$
Note how $2a_1mcos(\phi_1+\theta)$ does not depend on the parameters we seek. Therefore, if we reinject this in the original equation we get :
$$a_2^{2}+[2a_1sin(\phi_1+\theta)]a_2sin(\phi_2+\theta) = A^{2}-a_1^{2}-2a_1mcos(\phi_1+\theta)$$
If we name the right hand $\gamma$ and we set $\alpha = 2a_1sin(\phi_1+\theta)$ we get the nicer equation
$$a_2^{2}+\alpha a_2sin(\phi_2+\theta) = \gamma\tag{2}\label{eq2}$$
so we also have, by passing $a_2^{2}$ on the right side and squaring equation \eqref{eq2}:
$$\alpha^{2}a_2^{2}sin^{2}(\phi_2+\theta) = \gamma^{2}-2\gamma a_2^{2}+a_2^{4}$$
Now we remember we also have \eqref{eq1}. We multiply it and square it
$$\alpha^{2}a_2^{2}cos^{2}(\phi_2+\theta)=\alpha^{2}m^{2}$$
then, by summing both equations you eliminate the $\phi_2$ dependence. You finally get:
$$\alpha^{2}a_2^{2} = \gamma^{2}+\alpha^{2}m^{2}-2\gamma a_2^{2}+a_2^{4}$$
so
$$a_2^{4} -(\alpha^{2}+2\gamma)a_2^{2}+\gamma^{2}+m^{2}=0\tag{3}\label{eq3}$$
I believe \eqref{eq3} is called a bi squared equation. It means that you can solve it with $a_2^{2}$ as the unknown (you just have to find the roots of the polynomial $X^{2}-(\alpha^{2}+2\gamma)X+m^{2}+\gamma^{2}$) and then $a_2$ the generalized square roots of whatever you found for $a_2^{2}$ (with j or not, with + or -). You should get four roots and hope that only one makes sense.
Once you have $a_2$, for $\phi_2$ I suggest you use equations \eqref{eq2} and \eqref{eq1} :
$$ a_2sin(\phi_2+\theta) = \frac{\gamma-a_2^{2}}{\alpha}$$
$$a_2cos(\phi_2+\theta) = m$$
You ratio both equations and you solve with arctan but you can go with arccos on \eqref{eq1} or arcsine on \eqref{eq2}, it should depend on which hypothesis you make on $\phi_2$.
