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    Re: A triangle on the equator
    From: Frank Reed
    Date: 2025 Oct 9, 17:40 -0700

    David Pike, you wrote:
    "Using the little known and largely untested 'Bodger's Method', I get angle A = 53°13.12', which lies roughly half-way between the 3:4:5 method and Vincenty's formula method.  See photograph."

    Ha ha. You may want to explain to the crowd that your "duct tape" tinkering solution here is not intended as an actual method. It's an experiment! But you're absolutely "on to" an excellent approach to this puzzle. What you need is the size of a degree, both for latitude and longitude, at the equator. This is something that's available in standard tables, and there are well-known equations for it, too. Naturally the size of a degree scales to minutes of arc just fine. What do you get when you apply those lengths? Using the average length on a meridian of longitude does move the number in that right direction, and that's the hint that you're on the right track. :) 

    Frank Reed

       
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