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# angles and triangles watch

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1. 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
2. Need to use trigonometry.
The perpendicular is 'opposite' the angle
the horizontal is 'adjacent' to the angle

Tan(angle) = 1/100

angle = inv tan 1/100
angle = 0.57 degrees (to 2d.p.)
3. 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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