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18.1: Pre-Trigonometry

  • Page ID
    4721
  • Trigonometry Problem
    Figure \(\PageIndex{1}\): The basic problem of trigonometry.

    The page at http://www.phy6.org/stargaze/Strig1.htm describes the basic problem of trigonometry (drawing above): finding the distance to some far-away point C, given the directions at which C appears from the two ends of a measured baseline AB. This problem becomes somewhat simpler if:

    1. The baseline is perpendicular to the line from its middle to the object, so that \(\Delta ABC\) is symmetric. We will denote its equal sides as \(AB=BC=r\).
    2. The length \(c\) of the baseline \(AB\) is much smaller than \(r\). This means that the angle \(\alpha\) between \(AC\) and \(BC\) is small; that angle is known as the parallax of \(C\), as viewed from \(AB\).
    3. We do not ask for great accuracy, but are satisfied with an approximate value of the distance --- say, within 1%.
    Trigonometry Problem 2
    Figure \(\PageIndex{2}\): A simplified version of the problem (not to scale).

    The method presented here was already used by the ancient Greeks more than 2000 years ago. They knew that the length of a circle of radius \(r\) was \(2\pi r\), where \(\pi\) (a modern notation, not one of the Greeks, even though \(\pi\) is part of their alphabet) stands for a number a little larger than 3; approximately

    \(\pi\approx\,3.14159\ldots\)

    Note

    The Greek mathematician Archimedes derived \(\pi\) to about 4-figure accuracy, though he expressed it differently, since decimal fractions appeared in Europe only some 1000 years later.

    In this case (see Figure above), we can approximate\(\Delta ABC\) as a ‘slice’ of a much bigger circle; in this case, the length of the baseline is approximately equal to the length of the corresponding arc:

    \[c\,\approx\,c′\]

    There are 360 degrees in a circle and\(\alpha\) degrees in this particular arc; since 360 degrees corresponds to one circumference of arc length (\(2\pi r\)), \(\alpha\) degrees will correspond to an arclength of

    \[c′\,=\,\frac{\alpha}{360^o}\times 2\pi r\]

    Solving for \(r\) and plugging in \(c\,\approx\,c′\), we find

    \[r\,=\,\frac{360^o}{2\pi\alpha}\times c\]

    We have solved for \(r\) in terms of \(c\). For instance, if we know that \(\alpha\approx 6^o\) (we'll see why this is relevant later), \(2\pi\alpha\,=\,36^o\) and we get:

    \[r\,=\,10c′\]

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