Finding approximate latitude/longitude to the nearest minute (trigonometric)

Hello, this is my first post here on mathhelpforum. I hope to become an active member, helping and posting whatever I can/know to contribute.

I have been working on a problem that gives me three locations: A B and C. I have the latitude and longitude of A and B (approximate), and I have the degrees in which the directions of A B and C are pointing. By using a latitude and longitude to miles converter, I was able to find the length of side c, and since I had an opposite angle, I was able to use the Law of Sines to find the other sides, thus forming a triangle with all sides abc and angles ABC.

So I have the latitude and longitude of A (N46, W115.5), and the latitude and longitude of B (N48.16, W118.75).

A=35.83 degrees
B=54.16 degrees
C=90.01 degrees
a=125.20 miles
b=173.83 miles
c=213.87 miles

Given the information, what is the best way to approach finding the latitude and longitude of C to the nearest minute? I was guessing I could make a coordinate system, treating west as X and north as Y... I have those plotted... and do not know where to go from there.

Any help pushing me towards the right direction would be greatly appreciated.
See you around! (Evilgrin)

There's one important thing to remember.

1º longitude is 1º longitude, no matter where you are.

1º lattitude is substantially larger at the equator than 1º near the North or South Pole.

Generally, a good approximation can be achieved by attaching sin(longitude) to your lattitudes.

Let's see what you get.

Alright, so since I have the distances between abc, and since longitude is longitude no matter what, I began by using the distance between segment AB as a meter the find the miles / degree.

B = 118.75 long
A = 115.5 long

B - A = 3.25 difference.
c/3.25 = 65.80 miles/degree.

So since I have the distance between segment AC, I used the unit conversion with b to find the degree difference, and then subtract this difference from A's longitude?

b/65.80 = 2.64
115.5-2.64 = 112.86

Convert to the nearest second... 112° 51' 36"

Have I found the longitude of C correctly? ... or am I going about this the wrong way?

You will have to worry about geographical topography, but that is the idea. Try a few known cities and check it out.

Quote:

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Originally Posted by TKHunny

There's one important thing to remember.

1º longitude is 1º longitude, no matter where you are.

1º lattitude is substantially larger at the equator than 1º near the North or South Pole.

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TKHunny, I believe you have those reversed. Longitude goes east to west, latitude north and south. 1º of longitude, near the equator, is 1/360 of the circumference of the earth, about 21600/360= 60 nautical miles (in fact, the nautical mile is defined to be the length of 1 minute in longitude at the equator- it is remarkable that a nautical mile is so close to a statute mile). While close to the poles, the length of 1º longitude goes to 0.

Jeffh, what you really need is "spherical trigonometry" which used to be taught at the secondary level but is barely mentioned now.

This might help: http://www.boeing-727.com/Data/fly%20odds/distance.html

Quote:

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Generally, a good approximation can be achieved by attaching sin(longitude) to your lattitudes.

Let's see what you get.

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I am always doing that. Not sure why it won't stick. Sorry for reversal.

Note: If you see me in the cockpit of your airliner, please check to see who is navigating!