angle of view of a line segmentAngleOfViewOfALineSegment2007-10-04 11:28:372007-10-14 11:00:11Topicangleangle of view% this is the default PlanetMath preamble. as your knowledge
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Let $PQ$ be a line segment and $A$ a point not belonging to $PQ$. Let the magnitude of the angle $PAQ$ be $\alpha$. One says that the line segment $PQ$ {\em is seen from the point $A$ in an angle of $\alpha$}; one may also speak of the {\em angle of view} of $PQ$.
The locus of the points from which a given line segment $PQ$ is seen in an angle of $\alpha$ (with\, $0 < \alpha < 180^\circ$) consists of two congruent circular arcs having the line segment as the common chord and containing the circumferential angles equal to $\alpha$.
Especially, the locus of the points from which the line segment is seen in an angle of $90^\circ$ is the circle having the line segment as its diameter.
\begin{center}
\begin{pspicture}(-3,-4)(3,4)
\psline[linecolor=blue](-1.73,0)(1.73,0)
\rput[a](-2.1,-0.1){$P$}
\rput[a](2.1,-0.1){$Q$}
\psarc[linecolor=red](0,1){2}{-30}{210}
\psarc[linecolor=red](0,-1){2}{-210}{30}
\psline(-1.73,0)(-1.2,2.6)
\psline(1.73,0)(-1.2,2.6)
\psline(-1.73,0)(2,1)
\psline(1.73,0)(2,1)
\rput[a](-1.08,2.25){$\alpha$}
\rput[a](1.72,0.75){$\alpha$}
\psdots[linecolor=blue](-1.73,0)(1.73,0)
\psdots[linecolor=red](-1.2,2.6)(2,1)
\end{pspicture}
\end{center}
\textbf{Note.} The explementary arcs of the above mentioned two arcs form the locus of the points from which the segment $PQ$ is seen in the angle $180^\circ\!-\!\alpha$.