Example. Independent measurements of the angles of a triangle $x_{1}, x_{2}, x_{3}$ gave the results $x_{1}^{*}=31^{\circ}, x_{2}^{*}=62^{\circ}, x_{3}^{*}=89^{\circ}$. It is known that $x_{1}+x_{2}+x_{3}-180^{\circ}=0$. Use this equality to refine the values of $x_{1}^{*}, x_{2}^{*}, x_{3}^{*}$, assuming that the measurement errors of the angles $x_{1}, x_{2}, x_{3}$ are normally distributed with zero means and variances $\sigma_{1}^{2}=\sigma_{2}^{2}=\sigma_{3}^{2}=\left(0.1^{\circ}\right)^{2}$.
Solution. Let $\Delta x_{1}^{*}, \Delta x_{2}^{*}, \Delta x_{3}^{*}$ denote the errors contained in the numbers $x_{1}^{*}, x_{2}^{*}, x_{3}^{*}$.

1. We find $\Delta x_{1}^{*}, \Delta x_{2}^{*}, \Delta x_{3}^{*}$ from the condition

$$
\left\{\begin{array}{l}
\left(\frac{\Delta x_{1}^{*}}{0.1^{\circ}}\right)^{2}+\left(\frac{\Delta x_{2}^{*}}{0.1^{\circ}}\right)^{2}+\left(\frac{\Delta x_{3}^{*}}{0.1^{\circ}}\right)^{2} \longrightarrow \min \\
\left(x_{1}^{*}-\Delta x_{1}^{*}\right)+\left(x_{2}^{*}-\Delta x_{2}^{*}\right)+\left(x_{3}^{*}-\Delta x_{3}^{*}\right)-180^{\circ}=0
\end{array}\right.
$$

To do this, we use the method of Lagrange multipliers, i.e., we find the point of unconditional extremum of the function

$$
\begin{aligned}
& F\left(\Delta x_{1}^{*}, \Delta x_{2}^{*}, \Delta x_{3}^{*}, \lambda\right)=\left(\frac{\Delta x_{1}^{*}}{0.1^{\circ}}\right)^{2}+\left(\frac{\Delta x_{2}^{*}}{0.1^{\circ}}\right)^{2}+\left(\frac{\Delta x_{3}^{*}}{0.1^{\circ}}\right)^{2}+ \\
&+\lambda\left[\left(x_{1}^{*}-\Delta x_{1}^{*}\right)+\left(x_{2}^{*}-\Delta x_{2}^{*}\right)+\left(x_{3}^{*}-\Delta x_{3}^{*}\right)-180^{\circ}\right]
\end{aligned}
$$

a) The necessary condition for the extremum of the function $F$ is

$$
\left\{\begin{array}{l}
\frac{\partial F}{\partial \Delta x_{k}^{*}}=\frac{2}{\left(0.1^{\circ}\right)^{2}} \Delta x_{k}^{*}-\lambda=0, \quad k=1,2,3 \\
\frac{\partial F}{\partial \lambda}=x_{1}^{*}-\Delta x_{1}^{*}+x_{2}^{*}-\Delta x_{2}^{*}+x_{3}^{*}-\Delta x_{3}^{*}-180^{\circ}=0
\end{array}\right.
$$

b) We solve the system (2) and find $\Delta x_{1}^{*}, \Delta x_{2}^{*}, \Delta x_{3}^{*}$:

$$
\Delta x_{k}^{*}=\left(x_{1}^{*}+x_{2}^{*}+x_{3}^{*}\right) / 3=2^{\circ} / 3=40^{\prime}, \quad k=1,2,3 .
$$

2. We adjust the measurement results $x_{1}^{*}=31^{\circ}, x_{2}^{*}=62^{\circ}$, $x_{3}^{*}=89^{\circ}$, assuming

$$
\begin{aligned}
& \widetilde{x}_{1}=x_{1}^{*}-\Delta x_{1}^{*}=31^{\circ}-40^{\prime}=30^{\circ} 40^{\prime}, \\
& \widetilde{x}_{2}=x_{2}^{*}-\Delta x_{2}^{*}=62^{\circ}-40^{\prime}=61^{\circ} 40^{\prime} \\
& \widetilde{x}_{3}=x_{3}^{*}-\Delta x_{3}^{*}=89^{\circ}-40^{\prime}=88^{\circ} 20^{\prime}
\end{aligned}
$$

Answer. $x_{1}=30^{\circ} 40^{\prime}, x_{2}=61^{\circ} 40^{\prime}, x_{3}=88^{\circ} 20^{\prime}$.

Conditions of the Problem. Independent measurements of quantities $x_{1}, \ldots, x_{n}$ gave results $x_{1}^{*}, \ldots, x_{n}^{*}$. It is known that $f\left(x_{1}, \ldots, x_{n}\right)=0$. Use this equality to refine the values $x_{1}^{*}, \ldots, x_{n}^{*}$, assuming that the errors in measuring the quantities $x_{1}, \ldots, x_{n}$ are distributed according to normal laws with zero mathematical expectations and variances $\sigma_{1}^{2}, \ldots, \sigma_{n}^{2}$.

1. $x_{1}^{*}=17, x_{2}^{*}=84, x_{3}^{*}=83, x_{1}+x_{2}+x_{3}-180=0, \sigma_{1}^{2}=\sigma_{2}^{2}=\sigma_{3}^{2}=1$.
2. $x_{1}^{*}=52, x_{2}^{*}=47, x_{3}^{*}=82, x_{1}+x_{2}+x_{3}-180=0, \sigma_{1}^{2}=1, \sigma_{2}^{2}=2$, $\sigma_{3}^{2}=1.5$.
3. $x_{1}^{*}=0.27, x_{2}^{*}=0.15, x_{3}^{*}=0.57, x_{1}+x_{2}+x_{3}-1=0, \sigma_{1}^{2}=0.02$, $\sigma_{2}^{2}=0.03, \sigma_{3}^{2}=0.01$.
4. $x_{1}^{*}=0.89, x_{2}^{*}=0.46, x_{1}^{2}+x_{2}^{2}-1=0, \sigma_{1}^{2}=\sigma_{2}^{2}=0.01$.
5. $x_{1}^{*}=0.27, x_{2}^{*}=0.53, x_{3}^{*}=0.80, x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-1=0$, $\sigma_{1}^{2}=\sigma_{2}^{2}=\sigma_{3}^{2}=0.01$.
6. $x_{1}^{*}=0.89, x_{2}^{*}=0.46, x_{1}^{2}+x_{2}^{2}-1=0, \sigma_{1}^{2}=0.01, \sigma_{2}^{2}=0.02$.
7. $x_{1}^{*}=1.21, x_{2}^{*}=3.11, x_{3}^{*}=11.1, x_{3}-x_{1}^{2}-x_{2}^{2}=0$, $\sigma_{1}^{2}=\sigma_{2}^{2}=0.01, \sigma_{3}^{2}=0.02$.
8. $x_{1}^{*}=3.14, x_{2}^{*}=1.21, x_{3}^{*}=8.40, x_{3}-x_{1}^{2}+x_{2}^{2}=0$, $\sigma_{1}^{2}=\sigma_{2}^{2}=0.01, \sigma_{3}^{2}=0.02$.
9. $x_{1}^{*}=2.17, x_{2}^{*}=1.34, x_{3}^{*}=2.90, x_{3}-x_{1} x_{2}=0$, $\sigma_{1}^{2}=0.01, \sigma_{2}^{2}=0.02, \sigma_{3}^{2}=0.03$.
10. $x_{1}^{*}=3.12, x_{2}^{*}=2.87, x_{3}^{*}=8.95, x_{3}-x_{1} x_{2}=0$, $\sigma_{1}^{2}=0.02, \sigma_{2}^{2}=0.01, \sigma_{3}^{2}=0.03$.

Answers. 1. $\widetilde{x}_{1}=15 .(6), \widetilde{x}_{2}=82 .(6), \widetilde{x}_{3}=81 .(6) . \quad 2 . \widetilde{x}_{1}=51 .(7)$, $\widetilde{x}_{2}=46 .(5), \widetilde{x}_{3}=81 .(6)$. $3 . \widetilde{x}_{1}=0.27(3), \widetilde{x}_{2}=0.