# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

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Re: Another surveying problem
From: Lars Bergman
Date: 2021 Mar 2, 07:38 -0800

Mike, I've been looking into this problem, and I agree that you can't solve it by angles only. All angles except two, BAD and ABC, can be easily found, but you need one of those two to get the answer. By assigning CD = 1 and using the sine rule a couple of times you are able to find AX and BX, where X is the insersection of the diagonals. Then, knowing angle AXB, AB is found by the cosine rule and then angle BAD by the sine or cosine rule.

However, if you want to solve it "by hand", using log tables, the cosine rule is not very suitable. A better (at least I think so) method is to draw a perpendicular from B to the diagonal AD. If you call the intersection of the perpendicular and AD for X', then tan BAD = BX' / AX'.

BX' and XX' (and thence AX') are found from BX and angle BXX'.

Adding or subtracting two numbers when you have only the logarithms, and want the log of their sum or difference, is simplified if you have access to a table of Gaussian addition and subtraction logs. They are based on

log(a±b) = log(a·(1±b/a)) = log(a) + log(1±b/a) = log(a) + log(1±10^(log(b) - log(a)))

and when entered with (log(b) - log(a)) immediately (well, in my case interpolation was necessary) gives the term to be added to log(a) to get log(a±b). Unfortunately, those tables seems to be quite rare.

Enclosed you'll find my solution. I put it into Excel to get it readable, and as a check for errors. If you don't like logs starting with 9. ... (i.e. 10 added to a negative log), then it would have been better to use CD = 10 or even larger as you did. The result agrees with yours, 32°22'.

Lars

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