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A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Another surveying problem
From: David Pike
Date: 2021 Mar 6, 15:48 -0800
From: David Pike
Date: 2021 Mar 6, 15:48 -0800
Robert Bernecky you wrote: use the law of sines to set up 3 equations. With x = angle BAD:
sin(CAD + x) / BC = sin(ACB) / AB
sin(x) / BD = sin(ADB) / AB
sin(BCD) / BC = sin(BCD) / BD
The first two give sin(CAD + x) / sin(x) = sin(ACB) / sin(ADB) * (BC/BD). The third gives an expression for BC/BD.
Great solution Robert, without your Sin(CAD + x) lead, I’d still have been struggling. I ended up with the same expression Lars. I was a bit upset when I got 32.37°, but then I realised I was still in decimal. I changed 0.37° to minutes and joined the 32° 22’ club too. DaveP
