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    I last did maths about 15 years ago so I'm now puzzling over the following problem:

    I have a right-angled triangle with one side as 1m tall (the perpendicular). Another side (the horizontal) is 100m.

    How do I calculate the angle opposite the perpendicular?

    I believe it's the cosine of the base divided by the hypoteneuse but I'm not sure. Can anybody help me out on this one?

    R
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    Need to use trigonometry.
    The perpendicular is 'opposite' the angle
    the horizontal is 'adjacent' to the angle

    Tan(angle) = opposite/adjacent

    Tan(angle) = 1/100

    angle = inv tan 1/100
    angle = 0.57 degrees (to 2d.p.)

    Thanks. That's solved one half of the problem.

    Now for the other half and I'll explain what the problem was.

    I now have a perpendicular of 12mm, a base of unknown length and an angle opposite the base of 0.5792. Now I need to know the base length. I presume that'll be a simple tangent of .5792 multiplied by 0.012 (12mm in meters)?

    The problem was: somebody asked how long a camera lens had to be to take a decent picture of a family group, 100m away.

    I estimated that a family group would be 2m wide. Then obtain the angle at the lens to get the lens angle of view.

    Then I apply that angle to the image sensor in the camera (which is normally 1/2") to give the new base line. That base line is the focal length of lens required for the picture.

    I'm hoping that around 1200mm is correct.

    R
 
 
 
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