Intuitively, you'd want to give a larger weight to the more reliable predictor. So in the extreme case where one of the predictors is perfectly reliable (noise $\epsilon=0$), that predictor should have a weight of 1. Assigning a non-zero weight to the other predictor will not improve your estimate in this case, so the other weight should be 0. Here's a formal explanation.
Let ${\bf \Theta} =\big( \theta_1 \quad \theta_2 \big)^T$ denote the predictors (that satisfy the relations in your question) and $\Sigma=\begin{pmatrix} \sigma_1^2 & \rho\sigma_1\sigma_2 \\\rho\sigma_1\sigma_2 & \sigma_2^2 \end{pmatrix}$ be the covariance between noise in the predictors. Let ${\bf a}= \big( a_1 \quad a_2 \big)^T$ be the vector of weights on the predictors such that $\hat\theta={\bf a}^T{\bf \Theta}$ is an unbiased linear estimator of $\theta$ . The variance of this estimator must satisfy the Cramer-Rao bound, which (assuming Gaussian noise) for your case is:
\begin{align}
Var(\hat\theta) = {\bf a}^T\Sigma{\bf a} &\ge-\Big(E\Big[\frac{\partial^2lnP({\bf \Theta};\theta)}{\partial\theta^2}\Big]\Big)^{-1}\\
&\ge \Big(\frac{\partial\theta_1}{\partial\theta} \quad \frac{\partial\theta_2}{\partial\theta}\Big)\Sigma^{-1}\Big(\frac{\partial\theta_1}{\partial\theta} \quad \frac{\partial\theta_2}{\partial\theta}\Big)^T \\
& \ge \big(1 \quad 1\big)\Sigma^{-1}\big(1 \quad 1\big)^T
\end{align}
You can verify that the optimal value of ${\bf a}$ that minimizes the variance (i.e., the case for which equality holds) is: ${\bf a}^*=\frac{1}{\sigma_1^2+\sigma_2^2-2\rho\sigma_1\sigma_2}\begin{pmatrix} \sigma_2^2 - \rho\sigma_1\sigma_2 \\ \sigma_1^2 - \rho\sigma_1\sigma_2 \end{pmatrix}$.
From the above equation, you can see that if, for example, $\sigma_1=0$, then ${\bf a}^*=(1 \quad 0)^T$. You can play with different values of $\sigma_1, \sigma_2, $ and $\rho$ to see how they affect the optimal weights ${\bf a}^*$ and the quality of your estimator.