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en/14.1.3.md +76 −0
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+### Statement
+
+$14.1.3.$ [Insert the problem statement]
+
+### Solution
+
+![For problem $14.1.3$ |682x558, 31%](../../img/14.1.3/14.1.3_Observer.png)
+
+\documentclass[12pt,a4paper]{article}
+\usepackage[english]{babel}
+\usepackage{float}
+\usepackage{wrapfig}
+\usepackage{lmodern}
+\usepackage[T1]{fontenc}
+\usepackage[utf8]{inputenc}
+\usepackage{microtype}
+\usepackage{graphicx}
+\usepackage{booktabs}
+\usepackage{amsmath,amssymb}
+\usepackage{hyperref}
+\usepackage{csquotes}
+\usepackage{geometry}
+\usepackage{fancyhdr}
+\usepackage{subcaption}
+\usepackage{tikz}
+\usepackage{array}
+\usepackage{pgfplots}
+\usepackage{wrapfig}
+\usepackage{subcaption}
+
+
+\begin{document}
+
+\begin{center}
+ \Large \textbf{Statement}
+\end{center}
+
+At what angle to the horizon is a luminous object seen moving horizontally at
+a speed $\beta c$ at the moment when it is above the observer?
+
+\begin{center}
+ \Large \textbf{Solution}
+\end{center}
+
+The key idea of this problem is that light has a finite velocity $c$. To understand the solution, you may use the figure below:
+
+\begin{figure}[H]
+ \centering
+ \includegraphics[width=0.5\linewidth]{14.1.3_Observer.png}
+ \caption{The point $C$ is at the observer's head. The red vectors are the velocity vectors of the object at each position.}
+\end{figure}
+
+The light needs a time $t_1 = \frac{\overline{AC}}{c}$ to travel from $A$ to $C$. In this time, the object also moves. When the object reaches point $B$, the light reaches point $C$ at the same time. The time taken by the object is $t_2 = \frac{\overline{AB}}{\beta c}$. Since these times must be equal, we have:
+
+\begin{equation}
+ t_1 = t_2 \rightarrow \frac{\overline{AC}}{c} = \frac{\overline{AB}}{\beta c} \rightarrow \frac{\overline{AC}}{\overline{AB}} = \frac{1}{\beta}
+ \label{usingequation1}
+\end{equation}
+
+The angle between $\overline{AB}$ and $\overline{AC}$ can be calculated as:
+
+\begin{equation}
+ \cos\alpha = \frac{\overline{AB}}{\overline{AC}} \rightarrow \alpha = \cos^{-1} \left(\frac{\overline{AB}}{\overline{AC}}\right)
+\end{equation}
+
+Using equation \ref{usingequation1} we obtain:
+
+\begin{equation}
+ \alpha = \cos^{-1} \left(\frac{\overline{AB}}{\overline{AC}}\right) = \cos^{-1} (\beta)
+\end{equation}
+
+\end{document}
+
+#### Answer
+
+[Insert a concise answer or boxed result]