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angles and triangles watch


    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?


    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.

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